119317
The number of terms in the expansion of \((\mathrm{x}+\mathrm{y}+\mathrm{z})^{10}\) is
1 142
2 11
3 110
4 66
Explanation:
D The given expansion \((x+y+z)^{10}\) We know that, The number of term in the expansion of \(\left(\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\right.\) \(\left.\cdots+\mathrm{x}_{\mathrm{r}}\right)^{\mathrm{n}}={ }^{\mathrm{n}+\mathrm{r}-1} \mathrm{C}_{\mathrm{n}}\) Where, \(\mathrm{n}=\) exponent of the term \(r=\) number of terms Then, total number of terms - \({ }^{n+r-1} C_n ={ }^{10+3-1} C_{10}={ }^{12} C_{10}\) \(=\frac{12 !}{(12-10) ! 10 !}\) \(=\frac{12 \times 11 \times 10 !}{2 ! \times 10 !}\)So, total number of term \(=\frac{132}{2}=66\)
Karnataka CET-2020
Binomial Theorem and its Simple Application
119318
The number of terms in the expansion of \(\left(x^2+y^2\right)^{25}-\left(x^2-y^2\right)^{25}\) after simplification is
1 0
2 26
3 13
4 50
Explanation:
C Given, expansion is, \(\left(x^2+y^2\right)^{25}-\left(x^2-y^2\right)^{25}\) We know that, \((\mathrm{a}+\mathrm{b})^{\mathrm{n}}-(\mathrm{a}-\mathrm{b})^{\mathrm{n}}=\frac{\mathrm{n}+1}{2}\) Where \(\mathrm{n}\) is odd number. So, the number of terms in the given expansion \(\frac{\mathrm{n}+1}{2}=\frac{25+1}{2}=13\)
Karnataka CET-2019
Binomial Theorem and its Simple Application
119319
If \(\mathbf{n}\) is an odd positive integer and \(\left(1+x+x^2+\right.\) \(\left.x^3\right)^n=\sum_{r=0}^{3 n} a_r x^r\), then \(a_0-a_1+a_2-a_3+\ldots . .-a_{3 n}\) is
1 \(4^{\mathrm{n}}\)
2 1
3 -1
4 0
Explanation:
D Given, equation, \(\left(1+x+x^2+x^3\right)^n=\sum_{r=0}^{3 n} a_r x^r\) \(\text { Then, }\left(1+x+x^2+x^3\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{3 n} x^{3 n}\) \(\text { In the above expansion we put } x=-1 \text {, we get }-\) \((1-1+1-1)^n=a_0+a_1(-1)+a_2(-1)^2+\ldots .+a_{3 n}(-1)^{3 n}\) \(0=a_0-a_1+a_2-a_3+\ldots \ldots-a_{3 n}\) \(\text { So, } a_0-a_1+a_2-a_3+\ldots \ldots-a_{3 n}=0\)
Karnataka CET-2011
Binomial Theorem and its Simple Application
119320
The ninth term of the expansion \(\left(3 x-\frac{1}{2 x}\right)^8\) is
1 \(\frac{1}{512 \mathrm{x}^9}\)
2 \(\frac{-1}{512 \mathrm{x}^9}\)
3 \(\frac{-1}{256 x^8}\)
4 \(\frac{1}{256 \mathrm{x}^8}\)
Explanation:
D The given expression, \(\left(3 \mathrm{x}-\frac{1}{2 \mathrm{x}}\right)^8=\left[3 \mathrm{x}+\left(\frac{-1}{2 \mathrm{x}}\right)\right]^8\) Then, total number of terms \(=8+1=9\) We know that, The general term of the binomial expression - \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{a}^{\mathrm{r}} \cdot \mathrm{x}^{\mathrm{n}-\mathrm{r}}\) Here, \(\quad \mathrm{n}=8, \mathrm{x}=3 \mathrm{x}, \mathrm{a}=\frac{-1}{2 \mathrm{x}}\) For ninth term, \(r=8\) \(\therefore \quad \mathrm{T}_{8+1} ={ }^8 \mathrm{C}_{\mathrm{r}}(3 \mathrm{x})^{8-\mathrm{r}}\left(-\frac{1}{2 \mathrm{x}}\right)^{\mathrm{r}}\) \(\mathrm{T}_{8+1} ={ }^8 \mathrm{C}_8(3 \mathrm{x})^{8-8}\left(\frac{-1}{2 \mathrm{x}}\right)^8\)So, \(\quad \mathrm{T}_9=\frac{1}{256 \mathrm{x}^8}\)
Karnataka CET-2007
Binomial Theorem and its Simple Application
119321
If \(z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5\), then
119317
The number of terms in the expansion of \((\mathrm{x}+\mathrm{y}+\mathrm{z})^{10}\) is
1 142
2 11
3 110
4 66
Explanation:
D The given expansion \((x+y+z)^{10}\) We know that, The number of term in the expansion of \(\left(\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\right.\) \(\left.\cdots+\mathrm{x}_{\mathrm{r}}\right)^{\mathrm{n}}={ }^{\mathrm{n}+\mathrm{r}-1} \mathrm{C}_{\mathrm{n}}\) Where, \(\mathrm{n}=\) exponent of the term \(r=\) number of terms Then, total number of terms - \({ }^{n+r-1} C_n ={ }^{10+3-1} C_{10}={ }^{12} C_{10}\) \(=\frac{12 !}{(12-10) ! 10 !}\) \(=\frac{12 \times 11 \times 10 !}{2 ! \times 10 !}\)So, total number of term \(=\frac{132}{2}=66\)
Karnataka CET-2020
Binomial Theorem and its Simple Application
119318
The number of terms in the expansion of \(\left(x^2+y^2\right)^{25}-\left(x^2-y^2\right)^{25}\) after simplification is
1 0
2 26
3 13
4 50
Explanation:
C Given, expansion is, \(\left(x^2+y^2\right)^{25}-\left(x^2-y^2\right)^{25}\) We know that, \((\mathrm{a}+\mathrm{b})^{\mathrm{n}}-(\mathrm{a}-\mathrm{b})^{\mathrm{n}}=\frac{\mathrm{n}+1}{2}\) Where \(\mathrm{n}\) is odd number. So, the number of terms in the given expansion \(\frac{\mathrm{n}+1}{2}=\frac{25+1}{2}=13\)
Karnataka CET-2019
Binomial Theorem and its Simple Application
119319
If \(\mathbf{n}\) is an odd positive integer and \(\left(1+x+x^2+\right.\) \(\left.x^3\right)^n=\sum_{r=0}^{3 n} a_r x^r\), then \(a_0-a_1+a_2-a_3+\ldots . .-a_{3 n}\) is
1 \(4^{\mathrm{n}}\)
2 1
3 -1
4 0
Explanation:
D Given, equation, \(\left(1+x+x^2+x^3\right)^n=\sum_{r=0}^{3 n} a_r x^r\) \(\text { Then, }\left(1+x+x^2+x^3\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{3 n} x^{3 n}\) \(\text { In the above expansion we put } x=-1 \text {, we get }-\) \((1-1+1-1)^n=a_0+a_1(-1)+a_2(-1)^2+\ldots .+a_{3 n}(-1)^{3 n}\) \(0=a_0-a_1+a_2-a_3+\ldots \ldots-a_{3 n}\) \(\text { So, } a_0-a_1+a_2-a_3+\ldots \ldots-a_{3 n}=0\)
Karnataka CET-2011
Binomial Theorem and its Simple Application
119320
The ninth term of the expansion \(\left(3 x-\frac{1}{2 x}\right)^8\) is
1 \(\frac{1}{512 \mathrm{x}^9}\)
2 \(\frac{-1}{512 \mathrm{x}^9}\)
3 \(\frac{-1}{256 x^8}\)
4 \(\frac{1}{256 \mathrm{x}^8}\)
Explanation:
D The given expression, \(\left(3 \mathrm{x}-\frac{1}{2 \mathrm{x}}\right)^8=\left[3 \mathrm{x}+\left(\frac{-1}{2 \mathrm{x}}\right)\right]^8\) Then, total number of terms \(=8+1=9\) We know that, The general term of the binomial expression - \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{a}^{\mathrm{r}} \cdot \mathrm{x}^{\mathrm{n}-\mathrm{r}}\) Here, \(\quad \mathrm{n}=8, \mathrm{x}=3 \mathrm{x}, \mathrm{a}=\frac{-1}{2 \mathrm{x}}\) For ninth term, \(r=8\) \(\therefore \quad \mathrm{T}_{8+1} ={ }^8 \mathrm{C}_{\mathrm{r}}(3 \mathrm{x})^{8-\mathrm{r}}\left(-\frac{1}{2 \mathrm{x}}\right)^{\mathrm{r}}\) \(\mathrm{T}_{8+1} ={ }^8 \mathrm{C}_8(3 \mathrm{x})^{8-8}\left(\frac{-1}{2 \mathrm{x}}\right)^8\)So, \(\quad \mathrm{T}_9=\frac{1}{256 \mathrm{x}^8}\)
Karnataka CET-2007
Binomial Theorem and its Simple Application
119321
If \(z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5\), then
119317
The number of terms in the expansion of \((\mathrm{x}+\mathrm{y}+\mathrm{z})^{10}\) is
1 142
2 11
3 110
4 66
Explanation:
D The given expansion \((x+y+z)^{10}\) We know that, The number of term in the expansion of \(\left(\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\right.\) \(\left.\cdots+\mathrm{x}_{\mathrm{r}}\right)^{\mathrm{n}}={ }^{\mathrm{n}+\mathrm{r}-1} \mathrm{C}_{\mathrm{n}}\) Where, \(\mathrm{n}=\) exponent of the term \(r=\) number of terms Then, total number of terms - \({ }^{n+r-1} C_n ={ }^{10+3-1} C_{10}={ }^{12} C_{10}\) \(=\frac{12 !}{(12-10) ! 10 !}\) \(=\frac{12 \times 11 \times 10 !}{2 ! \times 10 !}\)So, total number of term \(=\frac{132}{2}=66\)
Karnataka CET-2020
Binomial Theorem and its Simple Application
119318
The number of terms in the expansion of \(\left(x^2+y^2\right)^{25}-\left(x^2-y^2\right)^{25}\) after simplification is
1 0
2 26
3 13
4 50
Explanation:
C Given, expansion is, \(\left(x^2+y^2\right)^{25}-\left(x^2-y^2\right)^{25}\) We know that, \((\mathrm{a}+\mathrm{b})^{\mathrm{n}}-(\mathrm{a}-\mathrm{b})^{\mathrm{n}}=\frac{\mathrm{n}+1}{2}\) Where \(\mathrm{n}\) is odd number. So, the number of terms in the given expansion \(\frac{\mathrm{n}+1}{2}=\frac{25+1}{2}=13\)
Karnataka CET-2019
Binomial Theorem and its Simple Application
119319
If \(\mathbf{n}\) is an odd positive integer and \(\left(1+x+x^2+\right.\) \(\left.x^3\right)^n=\sum_{r=0}^{3 n} a_r x^r\), then \(a_0-a_1+a_2-a_3+\ldots . .-a_{3 n}\) is
1 \(4^{\mathrm{n}}\)
2 1
3 -1
4 0
Explanation:
D Given, equation, \(\left(1+x+x^2+x^3\right)^n=\sum_{r=0}^{3 n} a_r x^r\) \(\text { Then, }\left(1+x+x^2+x^3\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{3 n} x^{3 n}\) \(\text { In the above expansion we put } x=-1 \text {, we get }-\) \((1-1+1-1)^n=a_0+a_1(-1)+a_2(-1)^2+\ldots .+a_{3 n}(-1)^{3 n}\) \(0=a_0-a_1+a_2-a_3+\ldots \ldots-a_{3 n}\) \(\text { So, } a_0-a_1+a_2-a_3+\ldots \ldots-a_{3 n}=0\)
Karnataka CET-2011
Binomial Theorem and its Simple Application
119320
The ninth term of the expansion \(\left(3 x-\frac{1}{2 x}\right)^8\) is
1 \(\frac{1}{512 \mathrm{x}^9}\)
2 \(\frac{-1}{512 \mathrm{x}^9}\)
3 \(\frac{-1}{256 x^8}\)
4 \(\frac{1}{256 \mathrm{x}^8}\)
Explanation:
D The given expression, \(\left(3 \mathrm{x}-\frac{1}{2 \mathrm{x}}\right)^8=\left[3 \mathrm{x}+\left(\frac{-1}{2 \mathrm{x}}\right)\right]^8\) Then, total number of terms \(=8+1=9\) We know that, The general term of the binomial expression - \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{a}^{\mathrm{r}} \cdot \mathrm{x}^{\mathrm{n}-\mathrm{r}}\) Here, \(\quad \mathrm{n}=8, \mathrm{x}=3 \mathrm{x}, \mathrm{a}=\frac{-1}{2 \mathrm{x}}\) For ninth term, \(r=8\) \(\therefore \quad \mathrm{T}_{8+1} ={ }^8 \mathrm{C}_{\mathrm{r}}(3 \mathrm{x})^{8-\mathrm{r}}\left(-\frac{1}{2 \mathrm{x}}\right)^{\mathrm{r}}\) \(\mathrm{T}_{8+1} ={ }^8 \mathrm{C}_8(3 \mathrm{x})^{8-8}\left(\frac{-1}{2 \mathrm{x}}\right)^8\)So, \(\quad \mathrm{T}_9=\frac{1}{256 \mathrm{x}^8}\)
Karnataka CET-2007
Binomial Theorem and its Simple Application
119321
If \(z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5\), then
NEET Test Series from KOTA - 10 Papers In MS WORD
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Binomial Theorem and its Simple Application
119317
The number of terms in the expansion of \((\mathrm{x}+\mathrm{y}+\mathrm{z})^{10}\) is
1 142
2 11
3 110
4 66
Explanation:
D The given expansion \((x+y+z)^{10}\) We know that, The number of term in the expansion of \(\left(\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\right.\) \(\left.\cdots+\mathrm{x}_{\mathrm{r}}\right)^{\mathrm{n}}={ }^{\mathrm{n}+\mathrm{r}-1} \mathrm{C}_{\mathrm{n}}\) Where, \(\mathrm{n}=\) exponent of the term \(r=\) number of terms Then, total number of terms - \({ }^{n+r-1} C_n ={ }^{10+3-1} C_{10}={ }^{12} C_{10}\) \(=\frac{12 !}{(12-10) ! 10 !}\) \(=\frac{12 \times 11 \times 10 !}{2 ! \times 10 !}\)So, total number of term \(=\frac{132}{2}=66\)
Karnataka CET-2020
Binomial Theorem and its Simple Application
119318
The number of terms in the expansion of \(\left(x^2+y^2\right)^{25}-\left(x^2-y^2\right)^{25}\) after simplification is
1 0
2 26
3 13
4 50
Explanation:
C Given, expansion is, \(\left(x^2+y^2\right)^{25}-\left(x^2-y^2\right)^{25}\) We know that, \((\mathrm{a}+\mathrm{b})^{\mathrm{n}}-(\mathrm{a}-\mathrm{b})^{\mathrm{n}}=\frac{\mathrm{n}+1}{2}\) Where \(\mathrm{n}\) is odd number. So, the number of terms in the given expansion \(\frac{\mathrm{n}+1}{2}=\frac{25+1}{2}=13\)
Karnataka CET-2019
Binomial Theorem and its Simple Application
119319
If \(\mathbf{n}\) is an odd positive integer and \(\left(1+x+x^2+\right.\) \(\left.x^3\right)^n=\sum_{r=0}^{3 n} a_r x^r\), then \(a_0-a_1+a_2-a_3+\ldots . .-a_{3 n}\) is
1 \(4^{\mathrm{n}}\)
2 1
3 -1
4 0
Explanation:
D Given, equation, \(\left(1+x+x^2+x^3\right)^n=\sum_{r=0}^{3 n} a_r x^r\) \(\text { Then, }\left(1+x+x^2+x^3\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{3 n} x^{3 n}\) \(\text { In the above expansion we put } x=-1 \text {, we get }-\) \((1-1+1-1)^n=a_0+a_1(-1)+a_2(-1)^2+\ldots .+a_{3 n}(-1)^{3 n}\) \(0=a_0-a_1+a_2-a_3+\ldots \ldots-a_{3 n}\) \(\text { So, } a_0-a_1+a_2-a_3+\ldots \ldots-a_{3 n}=0\)
Karnataka CET-2011
Binomial Theorem and its Simple Application
119320
The ninth term of the expansion \(\left(3 x-\frac{1}{2 x}\right)^8\) is
1 \(\frac{1}{512 \mathrm{x}^9}\)
2 \(\frac{-1}{512 \mathrm{x}^9}\)
3 \(\frac{-1}{256 x^8}\)
4 \(\frac{1}{256 \mathrm{x}^8}\)
Explanation:
D The given expression, \(\left(3 \mathrm{x}-\frac{1}{2 \mathrm{x}}\right)^8=\left[3 \mathrm{x}+\left(\frac{-1}{2 \mathrm{x}}\right)\right]^8\) Then, total number of terms \(=8+1=9\) We know that, The general term of the binomial expression - \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{a}^{\mathrm{r}} \cdot \mathrm{x}^{\mathrm{n}-\mathrm{r}}\) Here, \(\quad \mathrm{n}=8, \mathrm{x}=3 \mathrm{x}, \mathrm{a}=\frac{-1}{2 \mathrm{x}}\) For ninth term, \(r=8\) \(\therefore \quad \mathrm{T}_{8+1} ={ }^8 \mathrm{C}_{\mathrm{r}}(3 \mathrm{x})^{8-\mathrm{r}}\left(-\frac{1}{2 \mathrm{x}}\right)^{\mathrm{r}}\) \(\mathrm{T}_{8+1} ={ }^8 \mathrm{C}_8(3 \mathrm{x})^{8-8}\left(\frac{-1}{2 \mathrm{x}}\right)^8\)So, \(\quad \mathrm{T}_9=\frac{1}{256 \mathrm{x}^8}\)
Karnataka CET-2007
Binomial Theorem and its Simple Application
119321
If \(z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5\), then
119317
The number of terms in the expansion of \((\mathrm{x}+\mathrm{y}+\mathrm{z})^{10}\) is
1 142
2 11
3 110
4 66
Explanation:
D The given expansion \((x+y+z)^{10}\) We know that, The number of term in the expansion of \(\left(\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\right.\) \(\left.\cdots+\mathrm{x}_{\mathrm{r}}\right)^{\mathrm{n}}={ }^{\mathrm{n}+\mathrm{r}-1} \mathrm{C}_{\mathrm{n}}\) Where, \(\mathrm{n}=\) exponent of the term \(r=\) number of terms Then, total number of terms - \({ }^{n+r-1} C_n ={ }^{10+3-1} C_{10}={ }^{12} C_{10}\) \(=\frac{12 !}{(12-10) ! 10 !}\) \(=\frac{12 \times 11 \times 10 !}{2 ! \times 10 !}\)So, total number of term \(=\frac{132}{2}=66\)
Karnataka CET-2020
Binomial Theorem and its Simple Application
119318
The number of terms in the expansion of \(\left(x^2+y^2\right)^{25}-\left(x^2-y^2\right)^{25}\) after simplification is
1 0
2 26
3 13
4 50
Explanation:
C Given, expansion is, \(\left(x^2+y^2\right)^{25}-\left(x^2-y^2\right)^{25}\) We know that, \((\mathrm{a}+\mathrm{b})^{\mathrm{n}}-(\mathrm{a}-\mathrm{b})^{\mathrm{n}}=\frac{\mathrm{n}+1}{2}\) Where \(\mathrm{n}\) is odd number. So, the number of terms in the given expansion \(\frac{\mathrm{n}+1}{2}=\frac{25+1}{2}=13\)
Karnataka CET-2019
Binomial Theorem and its Simple Application
119319
If \(\mathbf{n}\) is an odd positive integer and \(\left(1+x+x^2+\right.\) \(\left.x^3\right)^n=\sum_{r=0}^{3 n} a_r x^r\), then \(a_0-a_1+a_2-a_3+\ldots . .-a_{3 n}\) is
1 \(4^{\mathrm{n}}\)
2 1
3 -1
4 0
Explanation:
D Given, equation, \(\left(1+x+x^2+x^3\right)^n=\sum_{r=0}^{3 n} a_r x^r\) \(\text { Then, }\left(1+x+x^2+x^3\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{3 n} x^{3 n}\) \(\text { In the above expansion we put } x=-1 \text {, we get }-\) \((1-1+1-1)^n=a_0+a_1(-1)+a_2(-1)^2+\ldots .+a_{3 n}(-1)^{3 n}\) \(0=a_0-a_1+a_2-a_3+\ldots \ldots-a_{3 n}\) \(\text { So, } a_0-a_1+a_2-a_3+\ldots \ldots-a_{3 n}=0\)
Karnataka CET-2011
Binomial Theorem and its Simple Application
119320
The ninth term of the expansion \(\left(3 x-\frac{1}{2 x}\right)^8\) is
1 \(\frac{1}{512 \mathrm{x}^9}\)
2 \(\frac{-1}{512 \mathrm{x}^9}\)
3 \(\frac{-1}{256 x^8}\)
4 \(\frac{1}{256 \mathrm{x}^8}\)
Explanation:
D The given expression, \(\left(3 \mathrm{x}-\frac{1}{2 \mathrm{x}}\right)^8=\left[3 \mathrm{x}+\left(\frac{-1}{2 \mathrm{x}}\right)\right]^8\) Then, total number of terms \(=8+1=9\) We know that, The general term of the binomial expression - \(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{a}^{\mathrm{r}} \cdot \mathrm{x}^{\mathrm{n}-\mathrm{r}}\) Here, \(\quad \mathrm{n}=8, \mathrm{x}=3 \mathrm{x}, \mathrm{a}=\frac{-1}{2 \mathrm{x}}\) For ninth term, \(r=8\) \(\therefore \quad \mathrm{T}_{8+1} ={ }^8 \mathrm{C}_{\mathrm{r}}(3 \mathrm{x})^{8-\mathrm{r}}\left(-\frac{1}{2 \mathrm{x}}\right)^{\mathrm{r}}\) \(\mathrm{T}_{8+1} ={ }^8 \mathrm{C}_8(3 \mathrm{x})^{8-8}\left(\frac{-1}{2 \mathrm{x}}\right)^8\)So, \(\quad \mathrm{T}_9=\frac{1}{256 \mathrm{x}^8}\)
Karnataka CET-2007
Binomial Theorem and its Simple Application
119321
If \(z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5\), then