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)
\[{}^n{C_r} = \dfrac{{\left| \!{\underline {\,
  n \,}} \right. }}{{\left| \!{\underline {\,
  r \,}} \right. \;\left| \!{\underline {\,
  {n - r} \,}} \right. }}\] 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.