119553
If the coefficients of \(x^5\) and \(x^6\) in \(\left(2+\frac{x}{3}\right)^n\) are equal, then \(\mathbf{n}\) is
1 51
2 31
3 41
4 None of these
Explanation:
C The expansion \(\left(2+\frac{\mathrm{x}}{3}\right)^{\mathrm{n}}\) Then general term in the expansion \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} 2^{\mathrm{n}-\mathrm{r}}\left(\frac{\mathrm{x}}{3}\right)^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} 2^{\mathrm{n}-\mathrm{r}} 3^{-\mathrm{r}} \mathrm{x}^{\mathrm{r}}\) The coefficient of \(\mathrm{x}^5\) and \(\mathrm{x}^6\) are equal. \({ }^n C_6 2^{n-6} 3^{-6} ={ }^n C_5 2^{n-5} 3^{-5}\) \(\frac{{ }^n C_6}{{ }^n C_5} =\frac{2^{n-5}}{2^{n-6}} \times \frac{3^{-5}}{3^{-6}}\) \(\frac{{ }^n C_6}{{ }^n C_5} =2 \times 3\) \(\frac{{ }^n C_6}{{ }^n C_5} =6\) \(=\frac{\mathrm{n} !}{6 !(\mathrm{n}-6) !} \times \frac{5 !(\mathrm{n}-5) !}{\mathrm{n} !}=6\) \(\frac{\mathrm{n}-5}{6}=6\) \(\mathrm{n}-5=36\) \(\mathrm{n}=41\)
COMEDK-2014
Binomial Theorem and its Simple Application
119554
If in the expansion of \((1+P x)^n, n \in N\), the coefficient of \(x\) and \(x^2\) are 8 and 24 , then
1 \(\mathrm{n}=3, \mathrm{p}=2\)
2 \(\mathrm{n}=5, \mathrm{p}=3\)
3 \(\mathrm{n}=4, \mathrm{p}=3\)
4 \(\mathrm{n}=4, \mathrm{p}=2\)
Explanation:
D The given expansion ( \(1+\mathrm{Px})^{\mathrm{n}}\) General term \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{P}^{\mathrm{r}} \mathrm{x}^{\mathrm{r}}(1)^{\mathrm{n}-\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{P}^{\mathrm{r}} \mathrm{X}^{\mathrm{r}}\) Then, the coefficient of \(x={ }^n C_1 P=8\) \(\mathrm{np}=8\) And also the coefficient of \(x^2\) \({ }^n \mathrm{C}_2 \mathrm{P}^2=24\) \(\frac{\mathrm{n}(\mathrm{n}-1)}{2} \mathrm{P}^2=24\) From the equation (i), we get - \(\frac{n(n-1)}{2}\left(\frac{8}{n}\right)^2=24\) \(\frac{64(n-1)}{2 n}=24\) \(4(n-1)=3 n\) \(4 n-4=3 n\) \(n=4\) Form equation (i), we get - \(4 . \mathrm{P}=8\) \(\mathrm{P}=2\)
COMEDK-2015
Binomial Theorem and its Simple Application
119556
Given the positive integers \(r>1, n>2\) and the coefficients of \((3 r)^{\text {th }}\) and \((r+2)^{\text {th }}\) terms in the expansion of \((1+x)^{2 n}\) and equal, then \(n=\)
1 \(2 \mathrm{r}\)
2 \(2 \mathrm{r}-1\)
3 \(2 \mathrm{r}+1\)
4 \(3 \mathrm{r}\)
Explanation:
A The given expansion \((1+\mathrm{x})^{2 \mathrm{n}}\) \(\mathrm{T}_{3 \mathrm{r}}={ }^{2 \mathrm{n}} \mathrm{C}_{3 \mathrm{r}-1}(\mathrm{x})^{3 \mathrm{r}-1}(1)^{3 \mathrm{r}-1}\) \(={ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1}(\mathrm{x})^{3 \mathrm{r}-1}\) \(\mathrm{~T}_{\mathrm{r}+2}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1}(\mathrm{x})^{\mathrm{r}+1}\) \(\text { According to the question }\) \({ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\) \(3 \mathrm{r}-1+\mathrm{r}+1=2 \mathrm{n}\) \(4 \mathrm{r}=2 \mathrm{n}\) \(\mathrm{n}=2 \mathrm{r}\)
COMEDK-2017
Binomial Theorem and its Simple Application
119557
The greatest value of the term independent of \(x\) in the expansion of ( \(x \quad \sin\) \(\left.\alpha+\mathbf{x}^{-1} \cos \alpha\right)^{10}, \alpha \in R\), is
1 \(\frac{10 !}{2^5}\)
2 \(\frac{10 !}{(5 !)^2}\)
3 \(\frac{1}{2^5} \frac{10 !}{(5 !)^2}\)
4 None of these
Explanation:
C The given expansion \(\left(x \sin \alpha+x^{-1} \cos \alpha\right)^{10}\) \(=\left(\mathrm{x} \sin \alpha+\frac{\cos \alpha}{\mathrm{x}}\right)^{10}\) \(\therefore \quad \mathrm{T}_{\mathrm{r}+1} ={ }^{10} \mathrm{C}_{\mathrm{r}}(\mathrm{x} \sin \alpha)^{10-\mathrm{r}}\left(\frac{\cos \alpha}{\mathrm{x}}\right)^{\mathrm{r}}\) \(={ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{10-\mathrm{r}}(\sin \alpha)^{10-\mathrm{r}}(\cos \alpha)^{\mathrm{r}} \mathrm{x}^{-\mathrm{r}}\) Then independent term of \(x\) then the power of \(x\) \(10-2 \mathrm{r}=0\) \(2 \mathrm{r}=10\) \(\mathrm{r}=5\) \(\therefore \mathrm{T}_{5+1}={ }^{10} \mathrm{C}_5(\sin \alpha)^5(\cos \alpha)^5\) \(={ }^{10} \mathrm{C}_5\left(\frac{2 \sin \alpha \cos \alpha}{2}\right)^5\) \(=\frac{{ }^{10} \mathrm{C}_5(\sin 2 \alpha)^5}{2^5}\) For maximum \(\sin 2 \alpha=1\) \(=\frac{10 !}{(5 !)^2 2^5}\)
COMEDK-2018
Binomial Theorem and its Simple Application
119558
If \(x^m\) occurs in the expansion of \(\left(x+\frac{1}{x^2}\right)^{2 n}\) then the coefficient of \(\mathrm{x}^{\mathrm{m}}\) is
119553
If the coefficients of \(x^5\) and \(x^6\) in \(\left(2+\frac{x}{3}\right)^n\) are equal, then \(\mathbf{n}\) is
1 51
2 31
3 41
4 None of these
Explanation:
C The expansion \(\left(2+\frac{\mathrm{x}}{3}\right)^{\mathrm{n}}\) Then general term in the expansion \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} 2^{\mathrm{n}-\mathrm{r}}\left(\frac{\mathrm{x}}{3}\right)^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} 2^{\mathrm{n}-\mathrm{r}} 3^{-\mathrm{r}} \mathrm{x}^{\mathrm{r}}\) The coefficient of \(\mathrm{x}^5\) and \(\mathrm{x}^6\) are equal. \({ }^n C_6 2^{n-6} 3^{-6} ={ }^n C_5 2^{n-5} 3^{-5}\) \(\frac{{ }^n C_6}{{ }^n C_5} =\frac{2^{n-5}}{2^{n-6}} \times \frac{3^{-5}}{3^{-6}}\) \(\frac{{ }^n C_6}{{ }^n C_5} =2 \times 3\) \(\frac{{ }^n C_6}{{ }^n C_5} =6\) \(=\frac{\mathrm{n} !}{6 !(\mathrm{n}-6) !} \times \frac{5 !(\mathrm{n}-5) !}{\mathrm{n} !}=6\) \(\frac{\mathrm{n}-5}{6}=6\) \(\mathrm{n}-5=36\) \(\mathrm{n}=41\)
COMEDK-2014
Binomial Theorem and its Simple Application
119554
If in the expansion of \((1+P x)^n, n \in N\), the coefficient of \(x\) and \(x^2\) are 8 and 24 , then
1 \(\mathrm{n}=3, \mathrm{p}=2\)
2 \(\mathrm{n}=5, \mathrm{p}=3\)
3 \(\mathrm{n}=4, \mathrm{p}=3\)
4 \(\mathrm{n}=4, \mathrm{p}=2\)
Explanation:
D The given expansion ( \(1+\mathrm{Px})^{\mathrm{n}}\) General term \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{P}^{\mathrm{r}} \mathrm{x}^{\mathrm{r}}(1)^{\mathrm{n}-\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{P}^{\mathrm{r}} \mathrm{X}^{\mathrm{r}}\) Then, the coefficient of \(x={ }^n C_1 P=8\) \(\mathrm{np}=8\) And also the coefficient of \(x^2\) \({ }^n \mathrm{C}_2 \mathrm{P}^2=24\) \(\frac{\mathrm{n}(\mathrm{n}-1)}{2} \mathrm{P}^2=24\) From the equation (i), we get - \(\frac{n(n-1)}{2}\left(\frac{8}{n}\right)^2=24\) \(\frac{64(n-1)}{2 n}=24\) \(4(n-1)=3 n\) \(4 n-4=3 n\) \(n=4\) Form equation (i), we get - \(4 . \mathrm{P}=8\) \(\mathrm{P}=2\)
COMEDK-2015
Binomial Theorem and its Simple Application
119556
Given the positive integers \(r>1, n>2\) and the coefficients of \((3 r)^{\text {th }}\) and \((r+2)^{\text {th }}\) terms in the expansion of \((1+x)^{2 n}\) and equal, then \(n=\)
1 \(2 \mathrm{r}\)
2 \(2 \mathrm{r}-1\)
3 \(2 \mathrm{r}+1\)
4 \(3 \mathrm{r}\)
Explanation:
A The given expansion \((1+\mathrm{x})^{2 \mathrm{n}}\) \(\mathrm{T}_{3 \mathrm{r}}={ }^{2 \mathrm{n}} \mathrm{C}_{3 \mathrm{r}-1}(\mathrm{x})^{3 \mathrm{r}-1}(1)^{3 \mathrm{r}-1}\) \(={ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1}(\mathrm{x})^{3 \mathrm{r}-1}\) \(\mathrm{~T}_{\mathrm{r}+2}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1}(\mathrm{x})^{\mathrm{r}+1}\) \(\text { According to the question }\) \({ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\) \(3 \mathrm{r}-1+\mathrm{r}+1=2 \mathrm{n}\) \(4 \mathrm{r}=2 \mathrm{n}\) \(\mathrm{n}=2 \mathrm{r}\)
COMEDK-2017
Binomial Theorem and its Simple Application
119557
The greatest value of the term independent of \(x\) in the expansion of ( \(x \quad \sin\) \(\left.\alpha+\mathbf{x}^{-1} \cos \alpha\right)^{10}, \alpha \in R\), is
1 \(\frac{10 !}{2^5}\)
2 \(\frac{10 !}{(5 !)^2}\)
3 \(\frac{1}{2^5} \frac{10 !}{(5 !)^2}\)
4 None of these
Explanation:
C The given expansion \(\left(x \sin \alpha+x^{-1} \cos \alpha\right)^{10}\) \(=\left(\mathrm{x} \sin \alpha+\frac{\cos \alpha}{\mathrm{x}}\right)^{10}\) \(\therefore \quad \mathrm{T}_{\mathrm{r}+1} ={ }^{10} \mathrm{C}_{\mathrm{r}}(\mathrm{x} \sin \alpha)^{10-\mathrm{r}}\left(\frac{\cos \alpha}{\mathrm{x}}\right)^{\mathrm{r}}\) \(={ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{10-\mathrm{r}}(\sin \alpha)^{10-\mathrm{r}}(\cos \alpha)^{\mathrm{r}} \mathrm{x}^{-\mathrm{r}}\) Then independent term of \(x\) then the power of \(x\) \(10-2 \mathrm{r}=0\) \(2 \mathrm{r}=10\) \(\mathrm{r}=5\) \(\therefore \mathrm{T}_{5+1}={ }^{10} \mathrm{C}_5(\sin \alpha)^5(\cos \alpha)^5\) \(={ }^{10} \mathrm{C}_5\left(\frac{2 \sin \alpha \cos \alpha}{2}\right)^5\) \(=\frac{{ }^{10} \mathrm{C}_5(\sin 2 \alpha)^5}{2^5}\) For maximum \(\sin 2 \alpha=1\) \(=\frac{10 !}{(5 !)^2 2^5}\)
COMEDK-2018
Binomial Theorem and its Simple Application
119558
If \(x^m\) occurs in the expansion of \(\left(x+\frac{1}{x^2}\right)^{2 n}\) then the coefficient of \(\mathrm{x}^{\mathrm{m}}\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Binomial Theorem and its Simple Application
119553
If the coefficients of \(x^5\) and \(x^6\) in \(\left(2+\frac{x}{3}\right)^n\) are equal, then \(\mathbf{n}\) is
1 51
2 31
3 41
4 None of these
Explanation:
C The expansion \(\left(2+\frac{\mathrm{x}}{3}\right)^{\mathrm{n}}\) Then general term in the expansion \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} 2^{\mathrm{n}-\mathrm{r}}\left(\frac{\mathrm{x}}{3}\right)^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} 2^{\mathrm{n}-\mathrm{r}} 3^{-\mathrm{r}} \mathrm{x}^{\mathrm{r}}\) The coefficient of \(\mathrm{x}^5\) and \(\mathrm{x}^6\) are equal. \({ }^n C_6 2^{n-6} 3^{-6} ={ }^n C_5 2^{n-5} 3^{-5}\) \(\frac{{ }^n C_6}{{ }^n C_5} =\frac{2^{n-5}}{2^{n-6}} \times \frac{3^{-5}}{3^{-6}}\) \(\frac{{ }^n C_6}{{ }^n C_5} =2 \times 3\) \(\frac{{ }^n C_6}{{ }^n C_5} =6\) \(=\frac{\mathrm{n} !}{6 !(\mathrm{n}-6) !} \times \frac{5 !(\mathrm{n}-5) !}{\mathrm{n} !}=6\) \(\frac{\mathrm{n}-5}{6}=6\) \(\mathrm{n}-5=36\) \(\mathrm{n}=41\)
COMEDK-2014
Binomial Theorem and its Simple Application
119554
If in the expansion of \((1+P x)^n, n \in N\), the coefficient of \(x\) and \(x^2\) are 8 and 24 , then
1 \(\mathrm{n}=3, \mathrm{p}=2\)
2 \(\mathrm{n}=5, \mathrm{p}=3\)
3 \(\mathrm{n}=4, \mathrm{p}=3\)
4 \(\mathrm{n}=4, \mathrm{p}=2\)
Explanation:
D The given expansion ( \(1+\mathrm{Px})^{\mathrm{n}}\) General term \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{P}^{\mathrm{r}} \mathrm{x}^{\mathrm{r}}(1)^{\mathrm{n}-\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{P}^{\mathrm{r}} \mathrm{X}^{\mathrm{r}}\) Then, the coefficient of \(x={ }^n C_1 P=8\) \(\mathrm{np}=8\) And also the coefficient of \(x^2\) \({ }^n \mathrm{C}_2 \mathrm{P}^2=24\) \(\frac{\mathrm{n}(\mathrm{n}-1)}{2} \mathrm{P}^2=24\) From the equation (i), we get - \(\frac{n(n-1)}{2}\left(\frac{8}{n}\right)^2=24\) \(\frac{64(n-1)}{2 n}=24\) \(4(n-1)=3 n\) \(4 n-4=3 n\) \(n=4\) Form equation (i), we get - \(4 . \mathrm{P}=8\) \(\mathrm{P}=2\)
COMEDK-2015
Binomial Theorem and its Simple Application
119556
Given the positive integers \(r>1, n>2\) and the coefficients of \((3 r)^{\text {th }}\) and \((r+2)^{\text {th }}\) terms in the expansion of \((1+x)^{2 n}\) and equal, then \(n=\)
1 \(2 \mathrm{r}\)
2 \(2 \mathrm{r}-1\)
3 \(2 \mathrm{r}+1\)
4 \(3 \mathrm{r}\)
Explanation:
A The given expansion \((1+\mathrm{x})^{2 \mathrm{n}}\) \(\mathrm{T}_{3 \mathrm{r}}={ }^{2 \mathrm{n}} \mathrm{C}_{3 \mathrm{r}-1}(\mathrm{x})^{3 \mathrm{r}-1}(1)^{3 \mathrm{r}-1}\) \(={ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1}(\mathrm{x})^{3 \mathrm{r}-1}\) \(\mathrm{~T}_{\mathrm{r}+2}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1}(\mathrm{x})^{\mathrm{r}+1}\) \(\text { According to the question }\) \({ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\) \(3 \mathrm{r}-1+\mathrm{r}+1=2 \mathrm{n}\) \(4 \mathrm{r}=2 \mathrm{n}\) \(\mathrm{n}=2 \mathrm{r}\)
COMEDK-2017
Binomial Theorem and its Simple Application
119557
The greatest value of the term independent of \(x\) in the expansion of ( \(x \quad \sin\) \(\left.\alpha+\mathbf{x}^{-1} \cos \alpha\right)^{10}, \alpha \in R\), is
1 \(\frac{10 !}{2^5}\)
2 \(\frac{10 !}{(5 !)^2}\)
3 \(\frac{1}{2^5} \frac{10 !}{(5 !)^2}\)
4 None of these
Explanation:
C The given expansion \(\left(x \sin \alpha+x^{-1} \cos \alpha\right)^{10}\) \(=\left(\mathrm{x} \sin \alpha+\frac{\cos \alpha}{\mathrm{x}}\right)^{10}\) \(\therefore \quad \mathrm{T}_{\mathrm{r}+1} ={ }^{10} \mathrm{C}_{\mathrm{r}}(\mathrm{x} \sin \alpha)^{10-\mathrm{r}}\left(\frac{\cos \alpha}{\mathrm{x}}\right)^{\mathrm{r}}\) \(={ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{10-\mathrm{r}}(\sin \alpha)^{10-\mathrm{r}}(\cos \alpha)^{\mathrm{r}} \mathrm{x}^{-\mathrm{r}}\) Then independent term of \(x\) then the power of \(x\) \(10-2 \mathrm{r}=0\) \(2 \mathrm{r}=10\) \(\mathrm{r}=5\) \(\therefore \mathrm{T}_{5+1}={ }^{10} \mathrm{C}_5(\sin \alpha)^5(\cos \alpha)^5\) \(={ }^{10} \mathrm{C}_5\left(\frac{2 \sin \alpha \cos \alpha}{2}\right)^5\) \(=\frac{{ }^{10} \mathrm{C}_5(\sin 2 \alpha)^5}{2^5}\) For maximum \(\sin 2 \alpha=1\) \(=\frac{10 !}{(5 !)^2 2^5}\)
COMEDK-2018
Binomial Theorem and its Simple Application
119558
If \(x^m\) occurs in the expansion of \(\left(x+\frac{1}{x^2}\right)^{2 n}\) then the coefficient of \(\mathrm{x}^{\mathrm{m}}\) is
119553
If the coefficients of \(x^5\) and \(x^6\) in \(\left(2+\frac{x}{3}\right)^n\) are equal, then \(\mathbf{n}\) is
1 51
2 31
3 41
4 None of these
Explanation:
C The expansion \(\left(2+\frac{\mathrm{x}}{3}\right)^{\mathrm{n}}\) Then general term in the expansion \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} 2^{\mathrm{n}-\mathrm{r}}\left(\frac{\mathrm{x}}{3}\right)^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} 2^{\mathrm{n}-\mathrm{r}} 3^{-\mathrm{r}} \mathrm{x}^{\mathrm{r}}\) The coefficient of \(\mathrm{x}^5\) and \(\mathrm{x}^6\) are equal. \({ }^n C_6 2^{n-6} 3^{-6} ={ }^n C_5 2^{n-5} 3^{-5}\) \(\frac{{ }^n C_6}{{ }^n C_5} =\frac{2^{n-5}}{2^{n-6}} \times \frac{3^{-5}}{3^{-6}}\) \(\frac{{ }^n C_6}{{ }^n C_5} =2 \times 3\) \(\frac{{ }^n C_6}{{ }^n C_5} =6\) \(=\frac{\mathrm{n} !}{6 !(\mathrm{n}-6) !} \times \frac{5 !(\mathrm{n}-5) !}{\mathrm{n} !}=6\) \(\frac{\mathrm{n}-5}{6}=6\) \(\mathrm{n}-5=36\) \(\mathrm{n}=41\)
COMEDK-2014
Binomial Theorem and its Simple Application
119554
If in the expansion of \((1+P x)^n, n \in N\), the coefficient of \(x\) and \(x^2\) are 8 and 24 , then
1 \(\mathrm{n}=3, \mathrm{p}=2\)
2 \(\mathrm{n}=5, \mathrm{p}=3\)
3 \(\mathrm{n}=4, \mathrm{p}=3\)
4 \(\mathrm{n}=4, \mathrm{p}=2\)
Explanation:
D The given expansion ( \(1+\mathrm{Px})^{\mathrm{n}}\) General term \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{P}^{\mathrm{r}} \mathrm{x}^{\mathrm{r}}(1)^{\mathrm{n}-\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{P}^{\mathrm{r}} \mathrm{X}^{\mathrm{r}}\) Then, the coefficient of \(x={ }^n C_1 P=8\) \(\mathrm{np}=8\) And also the coefficient of \(x^2\) \({ }^n \mathrm{C}_2 \mathrm{P}^2=24\) \(\frac{\mathrm{n}(\mathrm{n}-1)}{2} \mathrm{P}^2=24\) From the equation (i), we get - \(\frac{n(n-1)}{2}\left(\frac{8}{n}\right)^2=24\) \(\frac{64(n-1)}{2 n}=24\) \(4(n-1)=3 n\) \(4 n-4=3 n\) \(n=4\) Form equation (i), we get - \(4 . \mathrm{P}=8\) \(\mathrm{P}=2\)
COMEDK-2015
Binomial Theorem and its Simple Application
119556
Given the positive integers \(r>1, n>2\) and the coefficients of \((3 r)^{\text {th }}\) and \((r+2)^{\text {th }}\) terms in the expansion of \((1+x)^{2 n}\) and equal, then \(n=\)
1 \(2 \mathrm{r}\)
2 \(2 \mathrm{r}-1\)
3 \(2 \mathrm{r}+1\)
4 \(3 \mathrm{r}\)
Explanation:
A The given expansion \((1+\mathrm{x})^{2 \mathrm{n}}\) \(\mathrm{T}_{3 \mathrm{r}}={ }^{2 \mathrm{n}} \mathrm{C}_{3 \mathrm{r}-1}(\mathrm{x})^{3 \mathrm{r}-1}(1)^{3 \mathrm{r}-1}\) \(={ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1}(\mathrm{x})^{3 \mathrm{r}-1}\) \(\mathrm{~T}_{\mathrm{r}+2}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1}(\mathrm{x})^{\mathrm{r}+1}\) \(\text { According to the question }\) \({ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\) \(3 \mathrm{r}-1+\mathrm{r}+1=2 \mathrm{n}\) \(4 \mathrm{r}=2 \mathrm{n}\) \(\mathrm{n}=2 \mathrm{r}\)
COMEDK-2017
Binomial Theorem and its Simple Application
119557
The greatest value of the term independent of \(x\) in the expansion of ( \(x \quad \sin\) \(\left.\alpha+\mathbf{x}^{-1} \cos \alpha\right)^{10}, \alpha \in R\), is
1 \(\frac{10 !}{2^5}\)
2 \(\frac{10 !}{(5 !)^2}\)
3 \(\frac{1}{2^5} \frac{10 !}{(5 !)^2}\)
4 None of these
Explanation:
C The given expansion \(\left(x \sin \alpha+x^{-1} \cos \alpha\right)^{10}\) \(=\left(\mathrm{x} \sin \alpha+\frac{\cos \alpha}{\mathrm{x}}\right)^{10}\) \(\therefore \quad \mathrm{T}_{\mathrm{r}+1} ={ }^{10} \mathrm{C}_{\mathrm{r}}(\mathrm{x} \sin \alpha)^{10-\mathrm{r}}\left(\frac{\cos \alpha}{\mathrm{x}}\right)^{\mathrm{r}}\) \(={ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{10-\mathrm{r}}(\sin \alpha)^{10-\mathrm{r}}(\cos \alpha)^{\mathrm{r}} \mathrm{x}^{-\mathrm{r}}\) Then independent term of \(x\) then the power of \(x\) \(10-2 \mathrm{r}=0\) \(2 \mathrm{r}=10\) \(\mathrm{r}=5\) \(\therefore \mathrm{T}_{5+1}={ }^{10} \mathrm{C}_5(\sin \alpha)^5(\cos \alpha)^5\) \(={ }^{10} \mathrm{C}_5\left(\frac{2 \sin \alpha \cos \alpha}{2}\right)^5\) \(=\frac{{ }^{10} \mathrm{C}_5(\sin 2 \alpha)^5}{2^5}\) For maximum \(\sin 2 \alpha=1\) \(=\frac{10 !}{(5 !)^2 2^5}\)
COMEDK-2018
Binomial Theorem and its Simple Application
119558
If \(x^m\) occurs in the expansion of \(\left(x+\frac{1}{x^2}\right)^{2 n}\) then the coefficient of \(\mathrm{x}^{\mathrm{m}}\) is
119553
If the coefficients of \(x^5\) and \(x^6\) in \(\left(2+\frac{x}{3}\right)^n\) are equal, then \(\mathbf{n}\) is
1 51
2 31
3 41
4 None of these
Explanation:
C The expansion \(\left(2+\frac{\mathrm{x}}{3}\right)^{\mathrm{n}}\) Then general term in the expansion \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} 2^{\mathrm{n}-\mathrm{r}}\left(\frac{\mathrm{x}}{3}\right)^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} 2^{\mathrm{n}-\mathrm{r}} 3^{-\mathrm{r}} \mathrm{x}^{\mathrm{r}}\) The coefficient of \(\mathrm{x}^5\) and \(\mathrm{x}^6\) are equal. \({ }^n C_6 2^{n-6} 3^{-6} ={ }^n C_5 2^{n-5} 3^{-5}\) \(\frac{{ }^n C_6}{{ }^n C_5} =\frac{2^{n-5}}{2^{n-6}} \times \frac{3^{-5}}{3^{-6}}\) \(\frac{{ }^n C_6}{{ }^n C_5} =2 \times 3\) \(\frac{{ }^n C_6}{{ }^n C_5} =6\) \(=\frac{\mathrm{n} !}{6 !(\mathrm{n}-6) !} \times \frac{5 !(\mathrm{n}-5) !}{\mathrm{n} !}=6\) \(\frac{\mathrm{n}-5}{6}=6\) \(\mathrm{n}-5=36\) \(\mathrm{n}=41\)
COMEDK-2014
Binomial Theorem and its Simple Application
119554
If in the expansion of \((1+P x)^n, n \in N\), the coefficient of \(x\) and \(x^2\) are 8 and 24 , then
1 \(\mathrm{n}=3, \mathrm{p}=2\)
2 \(\mathrm{n}=5, \mathrm{p}=3\)
3 \(\mathrm{n}=4, \mathrm{p}=3\)
4 \(\mathrm{n}=4, \mathrm{p}=2\)
Explanation:
D The given expansion ( \(1+\mathrm{Px})^{\mathrm{n}}\) General term \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{P}^{\mathrm{r}} \mathrm{x}^{\mathrm{r}}(1)^{\mathrm{n}-\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{P}^{\mathrm{r}} \mathrm{X}^{\mathrm{r}}\) Then, the coefficient of \(x={ }^n C_1 P=8\) \(\mathrm{np}=8\) And also the coefficient of \(x^2\) \({ }^n \mathrm{C}_2 \mathrm{P}^2=24\) \(\frac{\mathrm{n}(\mathrm{n}-1)}{2} \mathrm{P}^2=24\) From the equation (i), we get - \(\frac{n(n-1)}{2}\left(\frac{8}{n}\right)^2=24\) \(\frac{64(n-1)}{2 n}=24\) \(4(n-1)=3 n\) \(4 n-4=3 n\) \(n=4\) Form equation (i), we get - \(4 . \mathrm{P}=8\) \(\mathrm{P}=2\)
COMEDK-2015
Binomial Theorem and its Simple Application
119556
Given the positive integers \(r>1, n>2\) and the coefficients of \((3 r)^{\text {th }}\) and \((r+2)^{\text {th }}\) terms in the expansion of \((1+x)^{2 n}\) and equal, then \(n=\)
1 \(2 \mathrm{r}\)
2 \(2 \mathrm{r}-1\)
3 \(2 \mathrm{r}+1\)
4 \(3 \mathrm{r}\)
Explanation:
A The given expansion \((1+\mathrm{x})^{2 \mathrm{n}}\) \(\mathrm{T}_{3 \mathrm{r}}={ }^{2 \mathrm{n}} \mathrm{C}_{3 \mathrm{r}-1}(\mathrm{x})^{3 \mathrm{r}-1}(1)^{3 \mathrm{r}-1}\) \(={ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1}(\mathrm{x})^{3 \mathrm{r}-1}\) \(\mathrm{~T}_{\mathrm{r}+2}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1}(\mathrm{x})^{\mathrm{r}+1}\) \(\text { According to the question }\) \({ }^{2 n} \mathrm{C}_{3 \mathrm{r}-1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\) \(3 \mathrm{r}-1+\mathrm{r}+1=2 \mathrm{n}\) \(4 \mathrm{r}=2 \mathrm{n}\) \(\mathrm{n}=2 \mathrm{r}\)
COMEDK-2017
Binomial Theorem and its Simple Application
119557
The greatest value of the term independent of \(x\) in the expansion of ( \(x \quad \sin\) \(\left.\alpha+\mathbf{x}^{-1} \cos \alpha\right)^{10}, \alpha \in R\), is
1 \(\frac{10 !}{2^5}\)
2 \(\frac{10 !}{(5 !)^2}\)
3 \(\frac{1}{2^5} \frac{10 !}{(5 !)^2}\)
4 None of these
Explanation:
C The given expansion \(\left(x \sin \alpha+x^{-1} \cos \alpha\right)^{10}\) \(=\left(\mathrm{x} \sin \alpha+\frac{\cos \alpha}{\mathrm{x}}\right)^{10}\) \(\therefore \quad \mathrm{T}_{\mathrm{r}+1} ={ }^{10} \mathrm{C}_{\mathrm{r}}(\mathrm{x} \sin \alpha)^{10-\mathrm{r}}\left(\frac{\cos \alpha}{\mathrm{x}}\right)^{\mathrm{r}}\) \(={ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{10-\mathrm{r}}(\sin \alpha)^{10-\mathrm{r}}(\cos \alpha)^{\mathrm{r}} \mathrm{x}^{-\mathrm{r}}\) Then independent term of \(x\) then the power of \(x\) \(10-2 \mathrm{r}=0\) \(2 \mathrm{r}=10\) \(\mathrm{r}=5\) \(\therefore \mathrm{T}_{5+1}={ }^{10} \mathrm{C}_5(\sin \alpha)^5(\cos \alpha)^5\) \(={ }^{10} \mathrm{C}_5\left(\frac{2 \sin \alpha \cos \alpha}{2}\right)^5\) \(=\frac{{ }^{10} \mathrm{C}_5(\sin 2 \alpha)^5}{2^5}\) For maximum \(\sin 2 \alpha=1\) \(=\frac{10 !}{(5 !)^2 2^5}\)
COMEDK-2018
Binomial Theorem and its Simple Application
119558
If \(x^m\) occurs in the expansion of \(\left(x+\frac{1}{x^2}\right)^{2 n}\) then the coefficient of \(\mathrm{x}^{\mathrm{m}}\) is