119571
The value of \(r\) for which the coefficients of \((r-5)\) th and \((3 r+1)\) th terms in the expansion of \((1+x)^{12}\) are equal, is
1 4
2 9
3 12
4 None of these
Explanation:
D The given expansion \((1+\mathrm{x})^{12}\) \(\mathrm{~T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}(1)^{12-\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}-6+1}={ }^{12} \mathrm{C}_{\mathrm{r}-6 \mathrm{x}}(\mathrm{r}-6)\) \(\mathrm{T}_{3 \mathrm{r}+1}=\mathrm{T}_{3 \mathrm{r}+1}={ }^{12} \mathrm{C}_{3 \mathrm{r}}\) \(\text { According to question }\) \({ }^{12} \mathrm{C}_{\mathrm{r}-6 \mathrm{x}} \mathrm{x}-6={ }^{12} \mathrm{C}_{3 \mathrm{r}}\) \(\mathrm{r}-6=3 \mathrm{r}\) \(\mathrm{r}-6+3 \mathrm{r}=12\) \(4 \mathrm{r}-6=12\) \(4 \mathrm{r}=18\) \(\mathrm{r}=\frac{18}{4}=\frac{9}{2}\)
UPSEE-2011
Binomial Theorem and its Simple Application
119572
The value of \(\frac{1}{81^n}-\frac{10}{81^n}{ }^{2 n} C_1+\frac{10^2}{81^n} C_2-\frac{10^3}{81^n} \cdot{ }^{2 n} C_3+\ldots+\frac{10^{2 n}}{81^n}\) is
119574
If \(n\) is even, then in the expansion of \(\left(1+\frac{x^2}{2 !}+\frac{x^4}{2 !}+\ldots\right)^2\), then the coefficient of \(x^n\) is
1 \(\frac{2^{\mathrm{n}}}{\mathrm{n} !}\)
2 \(\frac{2^{\mathrm{n}}-2}{\mathrm{n} !}\)
3 \(\frac{2^{\mathrm{n}-1}-1}{\mathrm{n} !}\)
4 \(\frac{2^{n-1}}{n !}\)
Explanation:
D The given expansion \(\left(1+\frac{x^2}{2 !}+\frac{x^4}{2 !}+\ldots\right)^2\) \(\left(=\frac{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}{2}\right)^2\) \(=\frac{\left(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}\right)^2}{4}\) \(\frac{1}{4}\left(\mathrm{e}^{2 \mathrm{x}}+\mathrm{e}^{-2 \mathrm{x}}+2\right)\) \(\frac{1}{4}\left\{2\left(1+\frac{(2 \mathrm{x})^2}{2 !}+\frac{(2 \mathrm{x})^4}{4 !}+\ldots\right)+2\right\}\) The coefficient of \(x^n=\frac{1}{2}\left\{\frac{2^n}{n !}\right\}\) \(=\frac{2^{\mathrm{n}}}{2(\mathrm{n} !)}=\frac{2^{\mathrm{n}-1}}{\mathrm{n} !}\)\(\therefore \quad\) Coefficient of \(\mathrm{x}^{\mathrm{n}}=\frac{2^{\mathrm{n}-1}}{\mathrm{n} !}\)
BCECE-2013
Binomial Theorem and its Simple Application
119575
The coefficient of \(x^{20}\) in the expansion of \(\left(1+3 x+3 x^2+x^3\right)^{20}\) is
1 \({ }^{60} \mathrm{C}_{40}\)
2 \({ }^{30} \mathrm{C}_{20}\)
3 \({ }^{15} \mathrm{C}_2\)
4 None of these
Explanation:
A The given expression \(\left(1+3 \mathrm{x}+3 \mathrm{x}^2+\mathrm{x}^3\right)^{20}\) We know that \((1+x)^3=(1)^3+(x)^3+3(x \cdot 1)(x+1)\) \(=1+x^3+3 x^2+3 x\) \(\left\{(1+x)^3\right\}^{20}=(1+x)^{60}\) The coefficient of \(x^{20}\) in the binomial expansion \({ }^{60} \mathrm{C}_{20}\) \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}}\) \({ }^{60} \mathrm{C}_{20}={ }^{60} \mathrm{C}_{60-20}={ }^{60} \mathrm{C}_{40}\)
119571
The value of \(r\) for which the coefficients of \((r-5)\) th and \((3 r+1)\) th terms in the expansion of \((1+x)^{12}\) are equal, is
1 4
2 9
3 12
4 None of these
Explanation:
D The given expansion \((1+\mathrm{x})^{12}\) \(\mathrm{~T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}(1)^{12-\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}-6+1}={ }^{12} \mathrm{C}_{\mathrm{r}-6 \mathrm{x}}(\mathrm{r}-6)\) \(\mathrm{T}_{3 \mathrm{r}+1}=\mathrm{T}_{3 \mathrm{r}+1}={ }^{12} \mathrm{C}_{3 \mathrm{r}}\) \(\text { According to question }\) \({ }^{12} \mathrm{C}_{\mathrm{r}-6 \mathrm{x}} \mathrm{x}-6={ }^{12} \mathrm{C}_{3 \mathrm{r}}\) \(\mathrm{r}-6=3 \mathrm{r}\) \(\mathrm{r}-6+3 \mathrm{r}=12\) \(4 \mathrm{r}-6=12\) \(4 \mathrm{r}=18\) \(\mathrm{r}=\frac{18}{4}=\frac{9}{2}\)
UPSEE-2011
Binomial Theorem and its Simple Application
119572
The value of \(\frac{1}{81^n}-\frac{10}{81^n}{ }^{2 n} C_1+\frac{10^2}{81^n} C_2-\frac{10^3}{81^n} \cdot{ }^{2 n} C_3+\ldots+\frac{10^{2 n}}{81^n}\) is
119574
If \(n\) is even, then in the expansion of \(\left(1+\frac{x^2}{2 !}+\frac{x^4}{2 !}+\ldots\right)^2\), then the coefficient of \(x^n\) is
1 \(\frac{2^{\mathrm{n}}}{\mathrm{n} !}\)
2 \(\frac{2^{\mathrm{n}}-2}{\mathrm{n} !}\)
3 \(\frac{2^{\mathrm{n}-1}-1}{\mathrm{n} !}\)
4 \(\frac{2^{n-1}}{n !}\)
Explanation:
D The given expansion \(\left(1+\frac{x^2}{2 !}+\frac{x^4}{2 !}+\ldots\right)^2\) \(\left(=\frac{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}{2}\right)^2\) \(=\frac{\left(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}\right)^2}{4}\) \(\frac{1}{4}\left(\mathrm{e}^{2 \mathrm{x}}+\mathrm{e}^{-2 \mathrm{x}}+2\right)\) \(\frac{1}{4}\left\{2\left(1+\frac{(2 \mathrm{x})^2}{2 !}+\frac{(2 \mathrm{x})^4}{4 !}+\ldots\right)+2\right\}\) The coefficient of \(x^n=\frac{1}{2}\left\{\frac{2^n}{n !}\right\}\) \(=\frac{2^{\mathrm{n}}}{2(\mathrm{n} !)}=\frac{2^{\mathrm{n}-1}}{\mathrm{n} !}\)\(\therefore \quad\) Coefficient of \(\mathrm{x}^{\mathrm{n}}=\frac{2^{\mathrm{n}-1}}{\mathrm{n} !}\)
BCECE-2013
Binomial Theorem and its Simple Application
119575
The coefficient of \(x^{20}\) in the expansion of \(\left(1+3 x+3 x^2+x^3\right)^{20}\) is
1 \({ }^{60} \mathrm{C}_{40}\)
2 \({ }^{30} \mathrm{C}_{20}\)
3 \({ }^{15} \mathrm{C}_2\)
4 None of these
Explanation:
A The given expression \(\left(1+3 \mathrm{x}+3 \mathrm{x}^2+\mathrm{x}^3\right)^{20}\) We know that \((1+x)^3=(1)^3+(x)^3+3(x \cdot 1)(x+1)\) \(=1+x^3+3 x^2+3 x\) \(\left\{(1+x)^3\right\}^{20}=(1+x)^{60}\) The coefficient of \(x^{20}\) in the binomial expansion \({ }^{60} \mathrm{C}_{20}\) \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}}\) \({ }^{60} \mathrm{C}_{20}={ }^{60} \mathrm{C}_{60-20}={ }^{60} \mathrm{C}_{40}\)
119571
The value of \(r\) for which the coefficients of \((r-5)\) th and \((3 r+1)\) th terms in the expansion of \((1+x)^{12}\) are equal, is
1 4
2 9
3 12
4 None of these
Explanation:
D The given expansion \((1+\mathrm{x})^{12}\) \(\mathrm{~T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}(1)^{12-\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}-6+1}={ }^{12} \mathrm{C}_{\mathrm{r}-6 \mathrm{x}}(\mathrm{r}-6)\) \(\mathrm{T}_{3 \mathrm{r}+1}=\mathrm{T}_{3 \mathrm{r}+1}={ }^{12} \mathrm{C}_{3 \mathrm{r}}\) \(\text { According to question }\) \({ }^{12} \mathrm{C}_{\mathrm{r}-6 \mathrm{x}} \mathrm{x}-6={ }^{12} \mathrm{C}_{3 \mathrm{r}}\) \(\mathrm{r}-6=3 \mathrm{r}\) \(\mathrm{r}-6+3 \mathrm{r}=12\) \(4 \mathrm{r}-6=12\) \(4 \mathrm{r}=18\) \(\mathrm{r}=\frac{18}{4}=\frac{9}{2}\)
UPSEE-2011
Binomial Theorem and its Simple Application
119572
The value of \(\frac{1}{81^n}-\frac{10}{81^n}{ }^{2 n} C_1+\frac{10^2}{81^n} C_2-\frac{10^3}{81^n} \cdot{ }^{2 n} C_3+\ldots+\frac{10^{2 n}}{81^n}\) is
119574
If \(n\) is even, then in the expansion of \(\left(1+\frac{x^2}{2 !}+\frac{x^4}{2 !}+\ldots\right)^2\), then the coefficient of \(x^n\) is
1 \(\frac{2^{\mathrm{n}}}{\mathrm{n} !}\)
2 \(\frac{2^{\mathrm{n}}-2}{\mathrm{n} !}\)
3 \(\frac{2^{\mathrm{n}-1}-1}{\mathrm{n} !}\)
4 \(\frac{2^{n-1}}{n !}\)
Explanation:
D The given expansion \(\left(1+\frac{x^2}{2 !}+\frac{x^4}{2 !}+\ldots\right)^2\) \(\left(=\frac{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}{2}\right)^2\) \(=\frac{\left(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}\right)^2}{4}\) \(\frac{1}{4}\left(\mathrm{e}^{2 \mathrm{x}}+\mathrm{e}^{-2 \mathrm{x}}+2\right)\) \(\frac{1}{4}\left\{2\left(1+\frac{(2 \mathrm{x})^2}{2 !}+\frac{(2 \mathrm{x})^4}{4 !}+\ldots\right)+2\right\}\) The coefficient of \(x^n=\frac{1}{2}\left\{\frac{2^n}{n !}\right\}\) \(=\frac{2^{\mathrm{n}}}{2(\mathrm{n} !)}=\frac{2^{\mathrm{n}-1}}{\mathrm{n} !}\)\(\therefore \quad\) Coefficient of \(\mathrm{x}^{\mathrm{n}}=\frac{2^{\mathrm{n}-1}}{\mathrm{n} !}\)
BCECE-2013
Binomial Theorem and its Simple Application
119575
The coefficient of \(x^{20}\) in the expansion of \(\left(1+3 x+3 x^2+x^3\right)^{20}\) is
1 \({ }^{60} \mathrm{C}_{40}\)
2 \({ }^{30} \mathrm{C}_{20}\)
3 \({ }^{15} \mathrm{C}_2\)
4 None of these
Explanation:
A The given expression \(\left(1+3 \mathrm{x}+3 \mathrm{x}^2+\mathrm{x}^3\right)^{20}\) We know that \((1+x)^3=(1)^3+(x)^3+3(x \cdot 1)(x+1)\) \(=1+x^3+3 x^2+3 x\) \(\left\{(1+x)^3\right\}^{20}=(1+x)^{60}\) The coefficient of \(x^{20}\) in the binomial expansion \({ }^{60} \mathrm{C}_{20}\) \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}}\) \({ }^{60} \mathrm{C}_{20}={ }^{60} \mathrm{C}_{60-20}={ }^{60} \mathrm{C}_{40}\)
119571
The value of \(r\) for which the coefficients of \((r-5)\) th and \((3 r+1)\) th terms in the expansion of \((1+x)^{12}\) are equal, is
1 4
2 9
3 12
4 None of these
Explanation:
D The given expansion \((1+\mathrm{x})^{12}\) \(\mathrm{~T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}(1)^{12-\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}-6+1}={ }^{12} \mathrm{C}_{\mathrm{r}-6 \mathrm{x}}(\mathrm{r}-6)\) \(\mathrm{T}_{3 \mathrm{r}+1}=\mathrm{T}_{3 \mathrm{r}+1}={ }^{12} \mathrm{C}_{3 \mathrm{r}}\) \(\text { According to question }\) \({ }^{12} \mathrm{C}_{\mathrm{r}-6 \mathrm{x}} \mathrm{x}-6={ }^{12} \mathrm{C}_{3 \mathrm{r}}\) \(\mathrm{r}-6=3 \mathrm{r}\) \(\mathrm{r}-6+3 \mathrm{r}=12\) \(4 \mathrm{r}-6=12\) \(4 \mathrm{r}=18\) \(\mathrm{r}=\frac{18}{4}=\frac{9}{2}\)
UPSEE-2011
Binomial Theorem and its Simple Application
119572
The value of \(\frac{1}{81^n}-\frac{10}{81^n}{ }^{2 n} C_1+\frac{10^2}{81^n} C_2-\frac{10^3}{81^n} \cdot{ }^{2 n} C_3+\ldots+\frac{10^{2 n}}{81^n}\) is
119574
If \(n\) is even, then in the expansion of \(\left(1+\frac{x^2}{2 !}+\frac{x^4}{2 !}+\ldots\right)^2\), then the coefficient of \(x^n\) is
1 \(\frac{2^{\mathrm{n}}}{\mathrm{n} !}\)
2 \(\frac{2^{\mathrm{n}}-2}{\mathrm{n} !}\)
3 \(\frac{2^{\mathrm{n}-1}-1}{\mathrm{n} !}\)
4 \(\frac{2^{n-1}}{n !}\)
Explanation:
D The given expansion \(\left(1+\frac{x^2}{2 !}+\frac{x^4}{2 !}+\ldots\right)^2\) \(\left(=\frac{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}{2}\right)^2\) \(=\frac{\left(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}\right)^2}{4}\) \(\frac{1}{4}\left(\mathrm{e}^{2 \mathrm{x}}+\mathrm{e}^{-2 \mathrm{x}}+2\right)\) \(\frac{1}{4}\left\{2\left(1+\frac{(2 \mathrm{x})^2}{2 !}+\frac{(2 \mathrm{x})^4}{4 !}+\ldots\right)+2\right\}\) The coefficient of \(x^n=\frac{1}{2}\left\{\frac{2^n}{n !}\right\}\) \(=\frac{2^{\mathrm{n}}}{2(\mathrm{n} !)}=\frac{2^{\mathrm{n}-1}}{\mathrm{n} !}\)\(\therefore \quad\) Coefficient of \(\mathrm{x}^{\mathrm{n}}=\frac{2^{\mathrm{n}-1}}{\mathrm{n} !}\)
BCECE-2013
Binomial Theorem and its Simple Application
119575
The coefficient of \(x^{20}\) in the expansion of \(\left(1+3 x+3 x^2+x^3\right)^{20}\) is
1 \({ }^{60} \mathrm{C}_{40}\)
2 \({ }^{30} \mathrm{C}_{20}\)
3 \({ }^{15} \mathrm{C}_2\)
4 None of these
Explanation:
A The given expression \(\left(1+3 \mathrm{x}+3 \mathrm{x}^2+\mathrm{x}^3\right)^{20}\) We know that \((1+x)^3=(1)^3+(x)^3+3(x \cdot 1)(x+1)\) \(=1+x^3+3 x^2+3 x\) \(\left\{(1+x)^3\right\}^{20}=(1+x)^{60}\) The coefficient of \(x^{20}\) in the binomial expansion \({ }^{60} \mathrm{C}_{20}\) \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}}\) \({ }^{60} \mathrm{C}_{20}={ }^{60} \mathrm{C}_{60-20}={ }^{60} \mathrm{C}_{40}\)
119571
The value of \(r\) for which the coefficients of \((r-5)\) th and \((3 r+1)\) th terms in the expansion of \((1+x)^{12}\) are equal, is
1 4
2 9
3 12
4 None of these
Explanation:
D The given expansion \((1+\mathrm{x})^{12}\) \(\mathrm{~T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}(1)^{12-\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}\) \(\mathrm{T}_{\mathrm{r}-6+1}={ }^{12} \mathrm{C}_{\mathrm{r}-6 \mathrm{x}}(\mathrm{r}-6)\) \(\mathrm{T}_{3 \mathrm{r}+1}=\mathrm{T}_{3 \mathrm{r}+1}={ }^{12} \mathrm{C}_{3 \mathrm{r}}\) \(\text { According to question }\) \({ }^{12} \mathrm{C}_{\mathrm{r}-6 \mathrm{x}} \mathrm{x}-6={ }^{12} \mathrm{C}_{3 \mathrm{r}}\) \(\mathrm{r}-6=3 \mathrm{r}\) \(\mathrm{r}-6+3 \mathrm{r}=12\) \(4 \mathrm{r}-6=12\) \(4 \mathrm{r}=18\) \(\mathrm{r}=\frac{18}{4}=\frac{9}{2}\)
UPSEE-2011
Binomial Theorem and its Simple Application
119572
The value of \(\frac{1}{81^n}-\frac{10}{81^n}{ }^{2 n} C_1+\frac{10^2}{81^n} C_2-\frac{10^3}{81^n} \cdot{ }^{2 n} C_3+\ldots+\frac{10^{2 n}}{81^n}\) is
119574
If \(n\) is even, then in the expansion of \(\left(1+\frac{x^2}{2 !}+\frac{x^4}{2 !}+\ldots\right)^2\), then the coefficient of \(x^n\) is
1 \(\frac{2^{\mathrm{n}}}{\mathrm{n} !}\)
2 \(\frac{2^{\mathrm{n}}-2}{\mathrm{n} !}\)
3 \(\frac{2^{\mathrm{n}-1}-1}{\mathrm{n} !}\)
4 \(\frac{2^{n-1}}{n !}\)
Explanation:
D The given expansion \(\left(1+\frac{x^2}{2 !}+\frac{x^4}{2 !}+\ldots\right)^2\) \(\left(=\frac{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}{2}\right)^2\) \(=\frac{\left(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}\right)^2}{4}\) \(\frac{1}{4}\left(\mathrm{e}^{2 \mathrm{x}}+\mathrm{e}^{-2 \mathrm{x}}+2\right)\) \(\frac{1}{4}\left\{2\left(1+\frac{(2 \mathrm{x})^2}{2 !}+\frac{(2 \mathrm{x})^4}{4 !}+\ldots\right)+2\right\}\) The coefficient of \(x^n=\frac{1}{2}\left\{\frac{2^n}{n !}\right\}\) \(=\frac{2^{\mathrm{n}}}{2(\mathrm{n} !)}=\frac{2^{\mathrm{n}-1}}{\mathrm{n} !}\)\(\therefore \quad\) Coefficient of \(\mathrm{x}^{\mathrm{n}}=\frac{2^{\mathrm{n}-1}}{\mathrm{n} !}\)
BCECE-2013
Binomial Theorem and its Simple Application
119575
The coefficient of \(x^{20}\) in the expansion of \(\left(1+3 x+3 x^2+x^3\right)^{20}\) is
1 \({ }^{60} \mathrm{C}_{40}\)
2 \({ }^{30} \mathrm{C}_{20}\)
3 \({ }^{15} \mathrm{C}_2\)
4 None of these
Explanation:
A The given expression \(\left(1+3 \mathrm{x}+3 \mathrm{x}^2+\mathrm{x}^3\right)^{20}\) We know that \((1+x)^3=(1)^3+(x)^3+3(x \cdot 1)(x+1)\) \(=1+x^3+3 x^2+3 x\) \(\left\{(1+x)^3\right\}^{20}=(1+x)^{60}\) The coefficient of \(x^{20}\) in the binomial expansion \({ }^{60} \mathrm{C}_{20}\) \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}}\) \({ }^{60} \mathrm{C}_{20}={ }^{60} \mathrm{C}_{60-20}={ }^{60} \mathrm{C}_{40}\)
