Question

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

• A. -4
• B. 6
• C. -8
• D. 10

Answer 1

Answer: (A)

$1+x^4+x^5=a_0+a_1(1+x)+a_2(1+x{)}^2+$
$a_3(1+x{)}^3+a_4(1+x{)}^4+a_5(1+x{)}^5$
$=a_0+a_1(1+x)+a_2\left(1+2x+x^2\right)+a_3(1+3x$
$+3x^2+x^3)+a_4\left(1+4x+6x^2+4x^3+x^4\right)+a_5(1$
$+5x+10x^2+10x^3+5x^4+x^5)$
So, Coeff. of $x^i$ in $LHS=$ Coeff. of $x^i$ on $RHS$
$i=5\Rightarrow 1=a_5\dots$(i)
$i=4\Rightarrow 1=a_4+5a_5=a_4+5$
$\Rightarrow a_4=-4\dots$(ii)
$i=3\Rightarrow 0=a_3+4a_4+10a_5$
$\Rightarrow a_3-16+10=0$
$\Rightarrow a_3=6\dots$ (iii)
$i=2\Rightarrow 0=a_2+3a_3+6a_4+10a_5$
$\Rightarrow a_2+18-24+10=0$
$\Rightarrow a_2=-4$
Put $x=-1$
$1=a_0$
Now differentiate w.r.t. $x$.
$4x^3+5x^4=a_1+2a_2(1+x)+3a_3(1+x{)}^2+\dots .$
Put $x=-1$
$\Rightarrow 1=a_1$
Again differentiate w.r.t. $x$
$12x^2+20x^3=2x{a}_2+6a_3(1+x)$
Put $x=-1$
$12-20=2a_2$
$\Rightarrow a_2=-4$