Question

The coefficient of $x$ in the expansion $(1+x)(1+2x)(1+3x)\cdots (1+100x)$ is also equal to

• A. ${100}^2-{99}^2+{98}^2-{97}^2+{96}^2-{95}^2+\cdots +2^2-1^2$
• B. The sum of all $101$ A.M.'s inserted between $1$ and $99$
• C. The sum of all $100$ A.M.'s inserted between $1$ and $99$
• D. The sum of first $20$ terms of the series $1+(1+3)+(1+3+5)+(1+3+5+7)+\cdots$

Answer 1

Answer: (A), (B)

The correct options are
A ${100}^2-{99}^2+{98}^2-{97}^2+{96}^2-{95}^2+\cdots +2^2-1^2$
B The sum of all $101$ A.M.'s inserted between $1$ and $99$

Coefficient of $x$ in the expansion $(1+x)(1+2x)(1+3x)\cdots (1+100x)$
$=1+2+3+\cdots +100=\frac{100\times 101}{2}=5050$

Now,
${100}^2-{99}^2+{98}^2-{97}^2+{96}^2-{95}^2+\dots +2^2-1^2=({100}^2-{99}^2)+({98}^2-{97}^2)+({96}^2-{95}^2)+\dots +(2^2-1^2)=(100+99)(100-99)+(98+97)(98-97)+\dots +(2+1)(2-1)=100+99+98+97+\dots +2+1=5050$

Now,
Let $n$ A.M.'s are inserted between $1$ and $99$, then the sum of them
$5050=\frac{n+2}{2}[1+99]-(1+99)\Rightarrow 5050=100[\frac{n+2}{2}-1]\Rightarrow 5050=50n\Rightarrow n=101$
Therefore $101$ A.M.'s are inserted.

Now, the sum of first $20$ terms of the series
$1+(1+3)+(1+3+5)+(1+3+5+7)+\cdots$
General term of the series is
$T_r=1+3+5+7\cdots (2r-1)=r^2\Rightarrow S=\sum _{r=1}^{20}{T}_r=\sum _{r=1}^{20}{r}^2\Rightarrow S=\frac{n(n+1)(2n+1)}{6}\Rightarrow S=\frac{20\times 21\times 41}{6}=2870$