Question

The value of $\sum\limits_{r = 1}^n {{{( - 1)}^{r + 1}}} \dfrac{{{}^n{C_r}}}{{r + 1}}$is equal to

Answer 1

Hint : First we have to know the binomial series expansion. Then find the expansion of ${\left( {1 - x} \right)^n}$. To get ${\left( {n + 1} \right)^{th}}$term in the above expansion we have integrate the above expansion both sides between $x = 0$ to $x = 1$( From given expansion $x$ is a function of $r$without extra constant term). Simplify the above result and we get the value of the given expansion.

Complete step-by-step answer :
An expression consisting of two terms, connected by positive or negative sign is called a binomial expression. For example, $2x - 3y$, $x + 1$, $\dfrac{1}{x} - \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}$,etc., are all binomial expressions.
Binomial theorem: If $a$ and $b$are real numbers and $n$ is a positive integer, then
${\left( {a + b} \right)^n} = {}^n{C_0}{a^n} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + - - - + {}^n{C_r}{a^{n - r}}{b^r} + - - - + {}^n{C_n}{b^n}$
\Rightarrow $$$${\left( {a + b} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{a^{n - r}}{b^r}}--(1)

n \,}} \right. }}{{\left| \\!{\underline {\, r \,}} \right. \;\left| \\!{\underline {\, {n - r} \,}} \right. }}$$ for $$0 \leqslant r \leqslant n$$ The general term or $${\left( {r + 1} \right)^{th}}$$ term in the expansion is given by $${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$$. Replace $$a$$ by $$1$$ and $$b$$ by $$ - x$$ $${\left( {1 - x} \right)^n} = \sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{}^n{C_r}{x^r}} $$--(2) Integrating on both sides of the equation (2) with respect to $$x$$ between the limits $$x = 0$$ and $$x = 1$$( because to get $${\left( {n + 1} \right)^{th}}$$term in the above expansion). $$\int\limits_0^1 {{{\left( {1 - x} \right)}^n}} dx = \int\limits_0^1 {\left( {\sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{}^n{C_r}{x^r}} } \right)} dx$$ Since integral and summation symbols can be interchanged (Because the summation is independent of $$x$$ and integral depends only on $$x$$). So we get $$\int\limits_0^1 {{{\left( {1 - x} \right)}^n}} dx = \sum\limits_{r = 0}^n {\left( {{{\left( { - 1} \right)}^r}{}^n{C_r}\int\limits_0^1 {{x^r}dx} } \right)} $$ After integrating, we get $$ \Rightarrow \left[ { - \dfrac{{{{\left( {1 - x} \right)}^{n + 1}}}}{{n + 1}}} \right]_0^1 = \sum\limits_{r = 0}^n {\left( {{{\left( { - 1} \right)}^r}{}^n{C_r}\left[ {\dfrac{{{x^{r + 1}}}}{{r + 1}}} \right]_{x = 0}^{x = 1}} \right)} = \sum\limits_{r = 0}^n {\left( {{{\left( { - 1} \right)}^r}\dfrac{{{}^n{C_r}}}{{r + 1}} - 0} \right)} $$ Simplifying above equation, we get $$ \Rightarrow 0 - \left[ { - \dfrac{1}{{n + 1}}} \right] = {\left( { - 1} \right)^0}\dfrac{{{}^n{C_0}}}{{0 + 1}} + \sum\limits_{r = 1}^n {\left( {{{\left( { - 1} \right)}^r}\dfrac{{{}^n{C_r}}}{{r + 1}}} \right)} $$ $$ \Rightarrow \dfrac{1}{{n + 1}} = 1 + \sum\limits_{r = 1}^n {\left( {{{\left( { - 1} \right)}^r}\dfrac{{{}^n{C_r}}}{{r + 1}}} \right)} $$ Since we want to get the general term in the form which contains $${( - 1)^{r + 1}}$$ so multiple $$ - 1$$ two times in the second term series of the RHS of the above equation. We get $$\dfrac{1}{{n + 1}} = 1 - \sum\limits_{r = 1}^n {\left( {{{\left( { - 1} \right)}^{r + 1}}\dfrac{{{}^n{C_r}}}{{r + 1}}} \right)} $$ Hence, rewriting above equation we get $$\sum\limits_{r = 1}^n {\left( {{{\left( { - 1} \right)}^{r + 1}}\dfrac{{{}^n{C_r}}}{{r + 1}}} \right)} = \dfrac{n}{{n + 1}}$$. **So, the correct answer is “$$ \dfrac{n}{{n + 1}}$$”.** **Note** : Note that the total number of terms in the binomial expansion of $${\left( {a + b} \right)^n}$$ is $$\left( {n + 1} \right)$$, i.e., one more than the exponent $$n$$. In any term the sum of the indices (exponents) of $$a$$and $$b$$ is equal to $$n$$. The coefficients in the expansion follow a certain pattern known as pascal’s triangle.