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

& \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{aligned}$$ Put a =2, we get $ \dfrac{1}{n+1}-\dfrac{{{\left( -1 \right)}^{n+1}}}{n+1}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{\left( -1 \right)}^{r}}\left( \dfrac{{{2}^{r+1}}}{r+1} \right)} $ Multiplying both sides by -1, we get $ \dfrac{{{\left( -1 \right)}^{n+1}}}{n+1}-\dfrac{1}{n+1}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{\left( -1 \right)}^{r+1}}\left( \dfrac{{{2}^{r+1}}}{r+1} \right)} $ Writing the summation term in expanded form, we get $ \dfrac{{{\left( -1 \right)}^{n+1}}}{n+1}-\dfrac{1}{n+1}=-{{2}^{n}}{{C}_{0}}+\left( \dfrac{{{2}^{2}}}{2} \right)\left( ^{n}{{C}_{1}} \right)-\cdots +{{\left( -1 \right)}^{n+1}}\left( \dfrac{{{2}^{n+1}}}{r+1} \right)\left( ^{n}{{C}_{r}} \right) $ Hence, we have When n is odd $ {{\left( -1 \right)}^{n}}=-1 $ Hence, we have $ {{\left( -1 \right)}^{n+1}}=1 $ Hence, we have $$-{{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.