Question

If $S=(1+x^2{)}^{100}+2x^2(1+x^2{)}^{99}+3x^4(1+x^2{)}^{98}+\cdots +101x^{200}$, then the coefficient of $x^{100}$ in $S$ is

• A. ${}^{100}{C}_{50}$
• B. ${}^{101}{C}_{50}$
• C. ${}^{102}{C}_{50}$
• D. ${}^{103}{C}_{50}$

Answer 1

The correct option is C ${}^{102}{C}_{50}$

Given : $S=(1+x^2{)}^{100}+2x^2(1+x^2{)}^{99}+3x^4(1+x^2{)}^{98}+\cdots +101x^{200}$
Let $a=(1+x^2),b=x^2$, then
$S=a^{100}+2b{a}^{99}+3b^2{a}^{98}+\cdots +101b^{100}\frac{bS}{a}=b{a}^{99}+2b^2{a}^{98}+\cdots +100b^{100}+101\frac{b^{101}}{a}$
Subtracting above two equations,
$(1-\frac{b}{a})S=a^{100}+b{a}^{99}+b^2{a}^{98}+\cdots +b^{100}-101b^{101}{a}^{-1}\Rightarrow S=a^{101}+b{a}^{100}+b^2{a}^{99}+\cdots +b^{100}a-101b^{101}(\because a-b=1)\Rightarrow S=\frac{a^{101}({\left(\frac{b}{a}\right)}^{101}-1)}{\frac{b}{a}-1}-101b^{101}\Rightarrow S=a^{102}-(a+101)b^{101}$
Putting $a$ and $b$, we get
$S=(1+x^2{)}^{102}-(102+x^2)x^{202}$

Coefficient of $x^{100}$ in $(1+x^2{)}^{102}-(102+x^2)x^{202}$
$=$ Coefficient of $x^{100}$ in $(1+x^2{)}^{102}$
$={}^{102}{C}_{50}$