Question

If $(1+ax+x^2{)}^n=a_0+a_{1}x+a_2{x}^2+..........+a_{2n}{x}^{2n}$, then the value of ${\sum }_{r=0}^{2n}(3r+1)a_r$ is

• A. $(a+2{)}^{n+1}$
• B. $(3n+1)(a+2{)}^n$
• C. $(3a+1)(a+2{)}^n$
• D. None of these

Answer 1

Answer: (B)

$(3n+1)(a+2{)}^n$

Substituting $x=1$ in the above binomial expression we get
$(a+2{)}^n=a_0+a_1+...a_{2n}$
$={\sum }_{r=0}^{2n}{a}_r$ ...(i)
Differentiating the given question we get
$n(1+ax+x^2{)}^{n-1}(2x+a)=a_1+2a_{2}x+...2n{x}^{2n-1}{a}_{2n}$
Substituting $x=1$ we get.
$n(a+2{)}^{n-1}(a+2)=a_1+2a_2+...2n{a}_{2n}$
$=n(a+2{)}^n$
$={\sum }_{r=0}^{2n}r{a}_r$
Hence $3n(a+2{)}^n={\sum }_{r=0}^{2n}3r{a}_r$ ...(ii)
Adding i and ii we get
$3n(a+2{)}^n+(a+2{)}^n={\sum }_{r=0}^{2n}3r{a}_r+{\sum }_{r=0}^{2n}r{a}_r$
$(3n+1)(a+2{)}^n={\sum }_{r=0}^{2n}(3r+1)a_r$