Question

Consider $(1+x{)}^{2n}+{(1+2x+x^2)}^n=\sum _{r=0}^{2n}{a}_r{x}^r,n\in N.$ If $\sum _{r=0}^{2n}{a}_r=f\left(n\right),$ then

• A. $\sum _{n=1}^{\infty }\frac{1}{f\left(n\right)}=\frac{1}{6}$
• B. $\sum _{n=1}^{\infty }\frac{1}{f\left(n\right)}=\frac{3}{8}$
• C. largest value of $p$ for which $f\left(5\right)$ is divisible by $2^p$ is $11$
• D. largest value of $p$ for which $f\left(5\right)$ is divisible by $2^p$ is $9$

Answer 1

Answer: (A), (C)

largest value of $p$ for which $f\left(5\right)$ is divisible by $2^p$ is $11$

$(1+x{)}^{2n}+{(1+2x+x^2)}^n=\sum _{r=0}^{2n}{a}_r{x}^r$
$\Rightarrow (1+x{)}^{2n}+(1+x{)}^{2n}=\sum _{r=0}^{2n}{a}_r{x}^r$
$\Rightarrow 2(1+x{)}^{2n}=\sum _{r=0}^{2n}{a}_r{x}^r$
Put $x=1,$ we get
$f\left(n\right)=\sum _{r=0}^{2n}{a}_r=2^{2n+1}\cdots \left(1\right)$

Now,
$\sum _{n=1}^{\infty }\frac{1}{f\left(n\right)}=\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\cdots$
$=\frac{\frac{1}{2^3}}{1-\frac{1}{4}}=\frac{1}{6}$

From $\left(1\right),f\left(5\right)=2^{11}$
Largest value of $p$ for which $f\left(5\right)$ is divisible by $2^p$ is $11.$