Question

If $C_0+C_1+C_2+\dots \dots .+C_n=128$ then, $C_0-\frac{C_1}{2}+\frac{C_2}{3}-\frac{C_3}{4}+\dots \dots .=$

• A. $0$
• B. $8$
• C. $\frac{1}{8}$
• D. $78$

Answer 1

The correct option is C $\frac{1}{8}$

We have,

$(1+a{)}^n=1+a{}^n{C}_1+a^2{}^n{C}_2...+a^n{}^n{C}_n$

Putting $a=1$ we get
$\Rightarrow$ $2^n=1+{}^n{C}_1+{}^n{C}_2...+{}^n{C}_n$
$\Rightarrow$ $2^n=128$ ...(given that the sum of the coefficients is $128$)
$\Rightarrow$ $n=7$

$(1-a{)}^n=1-a{}^n{C}_1+a^2{}^n{C}_2...+(-1{)}^n{a}^n{}^n{C}_n$

Integrating equation both sides with respect to $a$, we get
$-\frac{(1-a{)}^{n+1}}{n+1}+C=a-\frac{a^2{}^n{C}_1}{2}+\frac{a^3{}^n{C}_2}{3}...+(-1{)}^n\frac{a^{n+1}{}^n{C}_n}{n+1}$

Now RHS is zero for $a=0$

Hence,
$-\frac{(1-a{)}^{n+1}-1}{n+1}=a-\frac{a^2{}^n{C}_1}{2}+\frac{a^3{}^n{C}_2}{3}...+(-1{)}^n\frac{a^{n+1}{}^n{C}_n}{n+1}$

Substituting $a=1$ and $n=7$ we get,
$\frac{1}{8}=1-\frac{{}^7{C}_1}{2}+\frac{{}^7{C}_2}{3}...-\frac{{}^7{C}_7}{8}$