Question

If $C_0,C_1$, ..., $C_n$ are binomial coefficients in the expansion of $(1+x{)}^n$, then
$C_0+C_1.\frac{x}{2}+C_2.\frac{x^2}{3}+----+C_n.\frac{x^n}{n+1}=$

• A. $\frac{(1+x{)}^{n+1}-1}{(n+1)}$
• B. $\frac{(1+x{)}^{n+1}-1}{(n+1)x}$
• C. $\frac{(1+x{)}^{n+1}+1}{(n+1)x}$
• D. $\frac{(1+x{)}^{n+1}+1}{(n+1)}$

Answer 1

Answer: (B)

$\frac{(1+x{)}^{n+1}-1}{(n+1)x}$

$(1+a{)}^n=1+a{}^n{C}_1+a^2{}^n{C}_2...+a^n{}^n{C}_n$
Integrating both the sides with respect to $a$, we get.
$\frac{(1+a{)}^{n+1}}{n+1}+C=a+\frac{a^2{}^n{C}_1}{2}+\frac{a^3{}^n{C}_2}{3}...+\frac{a^{n+1}{}^n{C}_n}{n+1}$
Now RHS is zero for $a=0$
Hence
$a+\frac{a^2{}^n{C}_1}{2}+\frac{a^3{}^n{C}_2}{3}...+\frac{a^{n+1}{}^n{C}_n}{n+1}=\frac{(1+a{)}^{n+1}-1}{n+1}$
Substituting $a=x$, we get
$x+\frac{x^2{}^n{C}_1}{2}+\frac{x^3{}^n{C}_2}{3}...+\frac{x^{n+1}{}^n{C}_n}{n+1}=\frac{(1+x{)}^{n+1}-1}{n+1}$
Now dividing both the sides by $a$ factor of $x$, we get.
$1+\frac{x{}^n{C}_1}{2}+\frac{x^2{}^n{C}_2}{3}...+\frac{x^n{}^n{C}_n}{n+1}=\frac{(1+x{)}^{n+1}-1}{x(n+1)}$
Hence the answer is Option B