Question

Find the value of the expression $ -{{2}^{n}}{{C}_{0}}+\left( \dfrac{{{2}^{2}}}{2} \right)\left( ^{n}{{C}_{1}} \right)-\cdots +{{\left( -1 \right)}^{n+1}}\left( \dfrac{{{2}^{n+1}}}{r+1} \right)\left( ^{n}{{C}_{r}} \right) $
[a] 0 if n is even
[b] 0 if n is odd
[c] $ \dfrac{1}{n+1} $ if n is even
[d] $ \dfrac{1}{n+1} $ if n is odd.

Answer 1

Hint:Use the binomial theorem which states that the expansion of the binomial $ {{\left( x+y \right)}^{n}} $ is given by $ {{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}} $ . Replace x by 1 and y by -x . Integrate both sides within limits 0 and a. Finally, put a = 2 in the resulting expression and hence find the value of the given expression

Complete step-by-step answer:
We know that
 $ {{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}} $
Replacing x by 1 and y by -x, we get
 $ {{\left( 1-x \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{\left( -1 \right)}^{r}}{{x}^{r}}} $
Integrating both sides with respect to x with the limite x =0 and x= a, we get
 $ \int_{0}^{a}{{{\left( 1-x \right)}^{n}}}dx=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}}{{\left( -1 \right)}^{r}}\int_{0}^{a}{{{x}^{r}}dx} $
Hence, we have
\[\begin{align}
  & \left. -\dfrac{{{\left( 1-x \right)}^{n+1}}}{n+1} \right|_{0}^{a}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{\left( -1 \right)}^{r}}\left( \left. \dfrac{{{x}^{r+1}}}{r+1} \right|_{0}^{a} \right)} \\
 & \Rightarrow \dfrac{1}{n+1}-\dfrac{{{\left( 1-a \right)}^{n+1}}}{n+1}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{\left( -1 \right)}^{r}}\left( \dfrac{{{a}^{r+1}}}{r+1} \right)} \\
\end{align}\]
Put a =2, we get
 $ \dfrac{1}{n+1}-\dfrac{{{\left( -1 \right)}^{n+1}}}{n+1}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{\left( -1 \right)}^{r}}\left( \dfrac{{{2}^{r+1}}}{r+1} \right)} $
Multiplying both sides by -1, we get
 $ \dfrac{{{\left( -1 \right)}^{n+1}}}{n+1}-\dfrac{1}{n+1}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{\left( -1 \right)}^{r+1}}\left( \dfrac{{{2}^{r+1}}}{r+1} \right)} $
Writing the summation term in expanded form, we get
 $ \dfrac{{{\left( -1 \right)}^{n+1}}}{n+1}-\dfrac{1}{n+1}=-{{2}^{n}}{{C}_{0}}+\left( \dfrac{{{2}^{2}}}{2} \right)\left( ^{n}{{C}_{1}} \right)-\cdots +{{\left( -1 \right)}^{n+1}}\left( \dfrac{{{2}^{n+1}}}{r+1} \right)\left( ^{n}{{C}_{r}} \right) $
Hence, we have
When n is odd $ {{\left( -1 \right)}^{n}}=-1 $
Hence, we have
 $ {{\left( -1 \right)}^{n+1}}=1 $
Hence, we have
\[-{{2}^{n}}{{C}_{0}}+\left( \dfrac{{{2}^{2}}}{2} \right)\left( ^{n}{{C}_{1}} \right)-\cdots +{{\left( -1 \right)}^{n+1}}\left( \dfrac{{{2}^{n+1}}}{r+1} \right)\left( ^{n}{{C}_{r}} \right)=\dfrac{1}{n+1}-\dfrac{1}{n+1}=0\]
When n is even, we have $ {{\left( -1 \right)}^{n}}=1 $
Hence, we have
 $ {{\left( -1 \right)}^{n+1}}=-1 $
Hence, we have
 $ -{{2}^{n}}{{C}_{0}}+\left( \dfrac{{{2}^{2}}}{2} \right)\left( ^{n}{{C}_{1}} \right)-\cdots +{{\left( -1 \right)}^{n+1}}\left( \dfrac{{{2}^{n+1}}}{r+1} \right)\left( ^{n}{{C}_{r}} \right)=\dfrac{-1}{n+1}-\dfrac{1}{n+1}=\dfrac{-2}{n+1} $
Hence option [b] is the only correct answer.

Note: Verification:
We can verify the correctness of our solution by checking LHS = RHS for n = 1, 2
For n = 1, we have
 $ LHS=-2{{\times }^{1}}{{C}_{0}}+\dfrac{{{2}^{2}}}{2}{{\times }^{1}}{{C}_{1}}=0 $
Also since 1 is odd, we have RHS = 0
True for n = 1.
For n = 2, we have
 $ LHS=-2{{\times }^{2}}{{C}_{0}}+\dfrac{{{2}^{2}}}{2}{{\times }^{2}}{{C}_{1}}-\dfrac{{{2}^{3}}}{3}{{\times }^{2}}{{C}_{2}}=-2+4-\dfrac{8}{3}=\dfrac{-2}{3} $
Also, we have
 $ RHS=\dfrac{-2}{2+1}=\dfrac{-2}{3} $
Hence, we have LHS = RHS
Hence our solution is verified to be correct.
[2] A student can make many calculation mistakes, especially in the signs and powers while solving this problem. Hence each should be checked at least twice.