Question

If $ 1 + x^4 + x^5 = \sum\limits^{5}_{i =0} a_i$ $(1 + x)^i$ , for all $x$ in $R$, then $a_2$ is:

• A. -4
• B. 6
• C. -8
• D. 10

Answer 1

The Correct Option is A

$1+x^{4}+x^{5}=a_{0}+a_{1}(1+x)+a_{2}(1+x)^{2}+$
$a_{3}(1+x)^{3}+a_{4}(1+x)^{4}+a_{5}(1+x)^{5} $
$=a_{0}+a_{1}(1+x)+a_{2}\left(1+2 x+x^{2}\right)+a_{3}(1+3 x$
$\left.+3 x^{2}+x^{3}\right)+a_{4}\left(1+4 x+6 x^{2}+4 x^{3}+x^{4}\right)+a_{5}(1 $
$\left.+5 x+10 x^{2}+10 x^{3}+5 x^{4}+x^{5}\right)$
So, Coeff. of $x ^{ i }$ in $LHS =$ Coeff. of $x ^{ i }$ on $RHS$
$i =5 \Rightarrow 1= a _{5} \ldots$(i)
$i =4 \Rightarrow 1= a _{4}+5 a _{5}= a _{4}+5 $
$ \Rightarrow a _{4}=-4 \ldots$(ii)
$i =3 \Rightarrow 0= a _{3}+4 a _{4}+10 a _{5} $
$ \Rightarrow a _{3}-16+10=0 $
$ \Rightarrow a _{3}=6 \ldots$ (iii)
$i =2 \Rightarrow 0= a _{2}+3 a _{3}+6 a _{4}+10 a _{5} $
$ \Rightarrow a _{2}+18-24+10=0 $
$ \Rightarrow a _{2}=-4 $
Put $ x =-1 $
$1= a _{0} $
Now differentiate w.r.t. $x$.
$4 x^{3}+5 x^{4}=a_{1}+2 a_{2}(1+x)+3 a_{3}(1+x)^{2}+\ldots .$
Put $x=-1$
$\Rightarrow 1=a_{1}$
Again differentiate w.r.t. $x$
$12 x^{2}+20 x^{3}=2 x a_{2}+6 a_{3}(1+x)$
Put $x=-1$
$12 - 20 = 2a_2 $
$\Rightarrow a_2 = -4$