119390
If \((1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots .+(-1)^n C_n x^n\) , then \(\frac{\mathrm{C}_0}{2}-\frac{\mathrm{C}_1}{3}+\frac{\mathrm{C}_2}{4}-\frac{\mathrm{C}_3}{5}+\ldots .+(-1)^{\mathrm{n}} \frac{\mathrm{C}_{\mathrm{n}}}{\mathrm{n}+2}\) is
1 \(\frac{1}{\mathrm{n}(\mathrm{n}+1)}\)
2 \(\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}\)
3 \(\frac{1}{(n+1)(n+3)}\)
4 \(\frac{1}{\mathrm{n}(\mathrm{n}+3)}\)
5 \(\frac{1}{(n+2)(n+3)}\)
Explanation:
B Given, \((1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots+(-1)^n C_n x^n\) Both side multiply by \(\mathrm{x}\) \(x(1-x)^n=x_0-C_1 x^2+C_2 x^3-C_3 \cdot x^4+\ldots+(-1)^n\). \(\mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}+1}\) On integrating w.r.t. \(\mathrm{x}\) \({\left[-x \frac{(1-x)^{n+1}}{n+1}-\frac{(1-x)^{n+2}}{(n+1)(n+2)}\right]_0^1}\) \(=\left[\frac{x^2}{2} C_0-\frac{C_1 x^3}{3}+\frac{C_2 x^4}{4} \ldots . .(-1)^n \frac{{ }^n C_n x^{n+2}}{n+2}\right]_0^1\) \(=\frac{(1-0)^{n+2}}{(n+1)(n+2)}=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}+\ldots(-1)^n \frac{C_n}{n+2}\) \(=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+\ldots .+(-1)^n \frac{C_n}{n+2}\) \(=\frac{1}{(n+1)(n+2)}\)
Kerala CEE-2011
Binomial Theorem and its Simple Application
119391
If the expansion of \(\left(\frac{3 \sqrt{x}}{7}-\frac{5}{2 x \sqrt{x}}\right)^{13 n}\) contains a term independent of \(x\) in the \(14^{\text {th }}\) term, then \(n\) should be
1 10
2 5
3 6
4 4
5 11
Explanation:
D Given, The \(14^{\text {th }}\) term in expansion of \(\left(\frac{3 \sqrt{\mathrm{x}}}{7}-\frac{5}{2 \mathrm{x} \sqrt{\mathrm{x}}}\right)^{13 \mathrm{n}}\) \(\mathrm{T}_{14}={ }^{13 \mathrm{n}} \mathrm{C}_{13}\left(\frac{3}{7} \mathrm{x}^{\frac{1}{2}}\right)^{13 \mathrm{n}-13}(-1)^{13}\left(\frac{5}{2} \mathrm{x}^{-3 / 2}\right)^{13}\) \(={ }^{13} \mathrm{C}_{13}\left(\frac{3}{7}\right)^{13 \mathrm{n}-13}(-1)^{13}\left(\frac{5}{2}\right)^{13} \cdot \mathrm{x}^{\frac{13 \mathrm{n}-13}{2}-\frac{39}{2}}\) For this term to be independent of \(x\), we get - \(13 \mathrm{n} =52\) \(\mathrm{n} =4\)
Kerala CEE-2008
Binomial Theorem and its Simple Application
119392
If \(n>1\), then \((1+x)^n-n x-1\) is divisible by
1 \(x^2\)
2 \(x^3\)
3 \(\mathrm{x}^4\)
4 \(x^5\)
Explanation:
A Given, \( \mathrm{n}>1\) \((1+\mathrm{x})^{\mathrm{n}}-\mathrm{nx}-1\) \(\text { By the binomial expansion }\) \(\Rightarrow\left[1+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3+\ldots . .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\right]-\mathrm{nx}-1\) \(\left.\Rightarrow 1+\mathrm{nx}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3 \ldots . .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{x}^{\mathrm{n}}\right]-\mathrm{nx}-1\) \(\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3 \ldots . .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{x}^{\mathrm{n}}\) \(\text { Hence, it is divisible by } \mathrm{x}^2\)Hence, it is divisible by \(x^2\).
AMU-2007
Binomial Theorem and its Simple Application
119552
The term independent of \(x\) in expansion of \(\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}\) is
119390
If \((1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots .+(-1)^n C_n x^n\) , then \(\frac{\mathrm{C}_0}{2}-\frac{\mathrm{C}_1}{3}+\frac{\mathrm{C}_2}{4}-\frac{\mathrm{C}_3}{5}+\ldots .+(-1)^{\mathrm{n}} \frac{\mathrm{C}_{\mathrm{n}}}{\mathrm{n}+2}\) is
1 \(\frac{1}{\mathrm{n}(\mathrm{n}+1)}\)
2 \(\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}\)
3 \(\frac{1}{(n+1)(n+3)}\)
4 \(\frac{1}{\mathrm{n}(\mathrm{n}+3)}\)
5 \(\frac{1}{(n+2)(n+3)}\)
Explanation:
B Given, \((1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots+(-1)^n C_n x^n\) Both side multiply by \(\mathrm{x}\) \(x(1-x)^n=x_0-C_1 x^2+C_2 x^3-C_3 \cdot x^4+\ldots+(-1)^n\). \(\mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}+1}\) On integrating w.r.t. \(\mathrm{x}\) \({\left[-x \frac{(1-x)^{n+1}}{n+1}-\frac{(1-x)^{n+2}}{(n+1)(n+2)}\right]_0^1}\) \(=\left[\frac{x^2}{2} C_0-\frac{C_1 x^3}{3}+\frac{C_2 x^4}{4} \ldots . .(-1)^n \frac{{ }^n C_n x^{n+2}}{n+2}\right]_0^1\) \(=\frac{(1-0)^{n+2}}{(n+1)(n+2)}=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}+\ldots(-1)^n \frac{C_n}{n+2}\) \(=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+\ldots .+(-1)^n \frac{C_n}{n+2}\) \(=\frac{1}{(n+1)(n+2)}\)
Kerala CEE-2011
Binomial Theorem and its Simple Application
119391
If the expansion of \(\left(\frac{3 \sqrt{x}}{7}-\frac{5}{2 x \sqrt{x}}\right)^{13 n}\) contains a term independent of \(x\) in the \(14^{\text {th }}\) term, then \(n\) should be
1 10
2 5
3 6
4 4
5 11
Explanation:
D Given, The \(14^{\text {th }}\) term in expansion of \(\left(\frac{3 \sqrt{\mathrm{x}}}{7}-\frac{5}{2 \mathrm{x} \sqrt{\mathrm{x}}}\right)^{13 \mathrm{n}}\) \(\mathrm{T}_{14}={ }^{13 \mathrm{n}} \mathrm{C}_{13}\left(\frac{3}{7} \mathrm{x}^{\frac{1}{2}}\right)^{13 \mathrm{n}-13}(-1)^{13}\left(\frac{5}{2} \mathrm{x}^{-3 / 2}\right)^{13}\) \(={ }^{13} \mathrm{C}_{13}\left(\frac{3}{7}\right)^{13 \mathrm{n}-13}(-1)^{13}\left(\frac{5}{2}\right)^{13} \cdot \mathrm{x}^{\frac{13 \mathrm{n}-13}{2}-\frac{39}{2}}\) For this term to be independent of \(x\), we get - \(13 \mathrm{n} =52\) \(\mathrm{n} =4\)
Kerala CEE-2008
Binomial Theorem and its Simple Application
119392
If \(n>1\), then \((1+x)^n-n x-1\) is divisible by
1 \(x^2\)
2 \(x^3\)
3 \(\mathrm{x}^4\)
4 \(x^5\)
Explanation:
A Given, \( \mathrm{n}>1\) \((1+\mathrm{x})^{\mathrm{n}}-\mathrm{nx}-1\) \(\text { By the binomial expansion }\) \(\Rightarrow\left[1+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3+\ldots . .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\right]-\mathrm{nx}-1\) \(\left.\Rightarrow 1+\mathrm{nx}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3 \ldots . .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{x}^{\mathrm{n}}\right]-\mathrm{nx}-1\) \(\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3 \ldots . .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{x}^{\mathrm{n}}\) \(\text { Hence, it is divisible by } \mathrm{x}^2\)Hence, it is divisible by \(x^2\).
AMU-2007
Binomial Theorem and its Simple Application
119552
The term independent of \(x\) in expansion of \(\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}\) is
119390
If \((1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots .+(-1)^n C_n x^n\) , then \(\frac{\mathrm{C}_0}{2}-\frac{\mathrm{C}_1}{3}+\frac{\mathrm{C}_2}{4}-\frac{\mathrm{C}_3}{5}+\ldots .+(-1)^{\mathrm{n}} \frac{\mathrm{C}_{\mathrm{n}}}{\mathrm{n}+2}\) is
1 \(\frac{1}{\mathrm{n}(\mathrm{n}+1)}\)
2 \(\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}\)
3 \(\frac{1}{(n+1)(n+3)}\)
4 \(\frac{1}{\mathrm{n}(\mathrm{n}+3)}\)
5 \(\frac{1}{(n+2)(n+3)}\)
Explanation:
B Given, \((1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots+(-1)^n C_n x^n\) Both side multiply by \(\mathrm{x}\) \(x(1-x)^n=x_0-C_1 x^2+C_2 x^3-C_3 \cdot x^4+\ldots+(-1)^n\). \(\mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}+1}\) On integrating w.r.t. \(\mathrm{x}\) \({\left[-x \frac{(1-x)^{n+1}}{n+1}-\frac{(1-x)^{n+2}}{(n+1)(n+2)}\right]_0^1}\) \(=\left[\frac{x^2}{2} C_0-\frac{C_1 x^3}{3}+\frac{C_2 x^4}{4} \ldots . .(-1)^n \frac{{ }^n C_n x^{n+2}}{n+2}\right]_0^1\) \(=\frac{(1-0)^{n+2}}{(n+1)(n+2)}=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}+\ldots(-1)^n \frac{C_n}{n+2}\) \(=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+\ldots .+(-1)^n \frac{C_n}{n+2}\) \(=\frac{1}{(n+1)(n+2)}\)
Kerala CEE-2011
Binomial Theorem and its Simple Application
119391
If the expansion of \(\left(\frac{3 \sqrt{x}}{7}-\frac{5}{2 x \sqrt{x}}\right)^{13 n}\) contains a term independent of \(x\) in the \(14^{\text {th }}\) term, then \(n\) should be
1 10
2 5
3 6
4 4
5 11
Explanation:
D Given, The \(14^{\text {th }}\) term in expansion of \(\left(\frac{3 \sqrt{\mathrm{x}}}{7}-\frac{5}{2 \mathrm{x} \sqrt{\mathrm{x}}}\right)^{13 \mathrm{n}}\) \(\mathrm{T}_{14}={ }^{13 \mathrm{n}} \mathrm{C}_{13}\left(\frac{3}{7} \mathrm{x}^{\frac{1}{2}}\right)^{13 \mathrm{n}-13}(-1)^{13}\left(\frac{5}{2} \mathrm{x}^{-3 / 2}\right)^{13}\) \(={ }^{13} \mathrm{C}_{13}\left(\frac{3}{7}\right)^{13 \mathrm{n}-13}(-1)^{13}\left(\frac{5}{2}\right)^{13} \cdot \mathrm{x}^{\frac{13 \mathrm{n}-13}{2}-\frac{39}{2}}\) For this term to be independent of \(x\), we get - \(13 \mathrm{n} =52\) \(\mathrm{n} =4\)
Kerala CEE-2008
Binomial Theorem and its Simple Application
119392
If \(n>1\), then \((1+x)^n-n x-1\) is divisible by
1 \(x^2\)
2 \(x^3\)
3 \(\mathrm{x}^4\)
4 \(x^5\)
Explanation:
A Given, \( \mathrm{n}>1\) \((1+\mathrm{x})^{\mathrm{n}}-\mathrm{nx}-1\) \(\text { By the binomial expansion }\) \(\Rightarrow\left[1+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3+\ldots . .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\right]-\mathrm{nx}-1\) \(\left.\Rightarrow 1+\mathrm{nx}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3 \ldots . .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{x}^{\mathrm{n}}\right]-\mathrm{nx}-1\) \(\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3 \ldots . .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{x}^{\mathrm{n}}\) \(\text { Hence, it is divisible by } \mathrm{x}^2\)Hence, it is divisible by \(x^2\).
AMU-2007
Binomial Theorem and its Simple Application
119552
The term independent of \(x\) in expansion of \(\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Binomial Theorem and its Simple Application
119390
If \((1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots .+(-1)^n C_n x^n\) , then \(\frac{\mathrm{C}_0}{2}-\frac{\mathrm{C}_1}{3}+\frac{\mathrm{C}_2}{4}-\frac{\mathrm{C}_3}{5}+\ldots .+(-1)^{\mathrm{n}} \frac{\mathrm{C}_{\mathrm{n}}}{\mathrm{n}+2}\) is
1 \(\frac{1}{\mathrm{n}(\mathrm{n}+1)}\)
2 \(\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}\)
3 \(\frac{1}{(n+1)(n+3)}\)
4 \(\frac{1}{\mathrm{n}(\mathrm{n}+3)}\)
5 \(\frac{1}{(n+2)(n+3)}\)
Explanation:
B Given, \((1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots+(-1)^n C_n x^n\) Both side multiply by \(\mathrm{x}\) \(x(1-x)^n=x_0-C_1 x^2+C_2 x^3-C_3 \cdot x^4+\ldots+(-1)^n\). \(\mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}+1}\) On integrating w.r.t. \(\mathrm{x}\) \({\left[-x \frac{(1-x)^{n+1}}{n+1}-\frac{(1-x)^{n+2}}{(n+1)(n+2)}\right]_0^1}\) \(=\left[\frac{x^2}{2} C_0-\frac{C_1 x^3}{3}+\frac{C_2 x^4}{4} \ldots . .(-1)^n \frac{{ }^n C_n x^{n+2}}{n+2}\right]_0^1\) \(=\frac{(1-0)^{n+2}}{(n+1)(n+2)}=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}+\ldots(-1)^n \frac{C_n}{n+2}\) \(=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+\ldots .+(-1)^n \frac{C_n}{n+2}\) \(=\frac{1}{(n+1)(n+2)}\)
Kerala CEE-2011
Binomial Theorem and its Simple Application
119391
If the expansion of \(\left(\frac{3 \sqrt{x}}{7}-\frac{5}{2 x \sqrt{x}}\right)^{13 n}\) contains a term independent of \(x\) in the \(14^{\text {th }}\) term, then \(n\) should be
1 10
2 5
3 6
4 4
5 11
Explanation:
D Given, The \(14^{\text {th }}\) term in expansion of \(\left(\frac{3 \sqrt{\mathrm{x}}}{7}-\frac{5}{2 \mathrm{x} \sqrt{\mathrm{x}}}\right)^{13 \mathrm{n}}\) \(\mathrm{T}_{14}={ }^{13 \mathrm{n}} \mathrm{C}_{13}\left(\frac{3}{7} \mathrm{x}^{\frac{1}{2}}\right)^{13 \mathrm{n}-13}(-1)^{13}\left(\frac{5}{2} \mathrm{x}^{-3 / 2}\right)^{13}\) \(={ }^{13} \mathrm{C}_{13}\left(\frac{3}{7}\right)^{13 \mathrm{n}-13}(-1)^{13}\left(\frac{5}{2}\right)^{13} \cdot \mathrm{x}^{\frac{13 \mathrm{n}-13}{2}-\frac{39}{2}}\) For this term to be independent of \(x\), we get - \(13 \mathrm{n} =52\) \(\mathrm{n} =4\)
Kerala CEE-2008
Binomial Theorem and its Simple Application
119392
If \(n>1\), then \((1+x)^n-n x-1\) is divisible by
1 \(x^2\)
2 \(x^3\)
3 \(\mathrm{x}^4\)
4 \(x^5\)
Explanation:
A Given, \( \mathrm{n}>1\) \((1+\mathrm{x})^{\mathrm{n}}-\mathrm{nx}-1\) \(\text { By the binomial expansion }\) \(\Rightarrow\left[1+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3+\ldots . .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\right]-\mathrm{nx}-1\) \(\left.\Rightarrow 1+\mathrm{nx}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3 \ldots . .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{x}^{\mathrm{n}}\right]-\mathrm{nx}-1\) \(\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^3 \ldots . .{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{x}^{\mathrm{n}}\) \(\text { Hence, it is divisible by } \mathrm{x}^2\)Hence, it is divisible by \(x^2\).
AMU-2007
Binomial Theorem and its Simple Application
119552
The term independent of \(x\) in expansion of \(\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}\) is
