Sum to (n + 1) terms of series frac{{{C_0}}}{2} - frac{{{C₁}}}{3} + frac{{{C₂}}}{4}

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7.Binomial Theorem

normal

The sum to $(n + 1)$ terms of the series $\frac{{{C_0}}}{2} - \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} - \frac{{{C_3}}}{5} + ...$ is

$\frac{1}{{n + 1}}$

$\frac{1}{{n + 2}}$

$\frac{1}{{n\left( {n + 1} \right)}}$

none of these

Solution

We have

${(1 – x)^n} = {C_0} – {C_1}x + {C_2}{x^2} – {C_3}{x^3} + …….$

$ \Rightarrow \quad x{(1 – x)^n} = {C_0}x – {C_1}{x^2} + {C_2}{x^3} – {C_3}{x^4} + …….$

$\int\limits_0^1 {x{{\left( {1 – x} \right)}^n}} dx = \int\limits_0^1 {x{{\left( {1 – x} \right)}^n}} dx$

$ = \int\limits_0^1 {\left( {1 – t} \right){t^n}} \left( { – 1} \right)dt\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[{\rm{ Put }}1 – x = t]$

$ = \int\limits_0^1 {\left( {{t^n} – {t^{n + 1}}} \right)} dt = \left| {\frac{{{t^{n + 1}}}}{{n + 1}} – \frac{{{t^{n + 2}}}}{{n + 2}}} \right|_0^1$

${=\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}}$

Integrating $R.H.S$ of $( 1)$ we get $\left. {\left( {\frac{{{C_0}{x^2}}}{2} – \frac{{{C_1}{x^3}}}{3} + \frac{{{C_2}{x^4}}}{4} + ……} \right)} \right|_0^1$

$=\frac{C_{0}}{2}-\frac{C_{1}}{3}+\frac{C_{2}}{4}-\ldots$

Thus $\quad \frac{C_{0}}{2}-\frac{C_{1}}{3}+\frac{C_{2}}{4}-\ldots=\frac{1}{(n+1)(n+2)}$

Standard 11

Mathematics

If $\frac{{ }^{11} C_1}{2}+\frac{{ }^{11} C_2}{3}+\ldots . .+\frac{{ }^{11} C_9}{10}=\frac{n}{m}$ with $\operatorname{gcd}(n, m)=1$, then $n+m$ is equal to

hard

(JEE MAIN-2024)

If ${a_1},{a_2},{a_3},{a_4}$ are the coefficients of any four consecutive terms in the expansion of ${(1 + x)^n}$, then $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$ =

hard

(IIT-1975)

If ${(1 – x + {x^2})^n} = {a_0} + {a_1}x + {a_2}{x^2} + …. + {a_{2n}}{x^{2n}}$, then ${a_0} + {a_2} + {a_4} + …. + {a_{2n}} = $

hard

Let $a =$ Minimum $\{x^2 + 2x + 3, x \in R\}$ and $b = \mathop {\lim }\limits_{\theta \to 0} \frac{{1 – \cos \theta }}{{{\theta ^2}}}$ The value of $\sum\limits_{r = 0}^n {{a^r}.{b^{n – r}}} $ is

normal

${n^n}{\left( {\frac{{n + 1}}{2}} \right)^{2n}}$ is

medium

The sum to $(n + 1)$ terms of the series $\frac{{{C_0}}}{2} - \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} - \frac{{{C_3}}}{5} + ...$ is

Solution

Similar Questions

If $\frac{{ }^{11} C_1}{2}+\frac{{ }^{11} C_2}{3}+\ldots . .+\frac{{ }^{11} C_9}{10}=\frac{n}{m}$ with $\operatorname{gcd}(n, m)=1$, then $n+m$ is equal to

If ${a_1},{a_2},{a_3},{a_4}$ are the coefficients of any four consecutive terms in the expansion of ${(1 + x)^n}$, then $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$ =

If ${(1 – x + {x^2})^n} = {a_0} + {a_1}x + {a_2}{x^2} + …. + {a_{2n}}{x^{2n}}$, then ${a_0} + {a_2} + {a_4} + …. + {a_{2n}} = $

Let $a =$ Minimum $\{x^2 + 2x + 3, x \in R\}$ and $b = \mathop {\lim }\limits_{\theta \to 0} \frac{{1 – \cos \theta }}{{{\theta ^2}}}$ The value of $\sum\limits_{r = 0}^n {{a^r}.{b^{n – r}}} $ is

${n^n}{\left( {\frac{{n + 1}}{2}} \right)^{2n}}$ is

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