Question

Q.3) 1^2+2^2-3^2-4^2+5^2+6^2-7^2-8^2+......-100^2=??
PLZZ solve question no.3​

Answer 1

We group the terms in the series as follows:-

$\begin{aligned}\longrightarrow\ \ &\sf{1^2+2^2-3^2-4^2+5^2+6^2-7^2-8^2+\,\dots\,+98^2-99^2-100^2}\\\\=\ \ &\sf{1^2+(2^2-3^2)-(4^2-5^2)+(6^2-7^2)-\,\dots\,+(98^2-99^2)-100^2}\end{aligned}$

Since $\sf{a^2-b^2=(a-b)(a+b),}$

$\begin{aligned}\longrightarrow\ \ &\sf{1^2+2^2-3^2-4^2+5^2+6^2-7^2-8^2+\,\dots\,+98^2-99^2-100^2}\\\\=\ \ &\sf{1+(2-3)(2+3)-(4-5)(4+5)+(6-7)(6+7)-\,\dots}\\\\&+\sf{(98-99)(98+99)-10000}\end{aligned}$

$\begin{aligned}\longrightarrow\ \ &\sf{1^2+2^2-3^2-4^2+5^2+6^2-7^2-8^2+\,\dots\,+98^2-99^2-100^2}\\\\=\ \ &\sf{1+(-1)(2+3)-(-1)(4+5)+(-1)(6+7)-\,\dots\,+(-1)(98+99)-10000}\end{aligned}$

$\begin{aligned}\longrightarrow\ \ &\sf{1^2+2^2-3^2-4^2+5^2+6^2-7^2-8^2+\,\dots\,+98^2-99^2-100^2}\\\\=\ \ &\sf{1-(2+3)+(4+5)-(6+7)+\,\dots\,-(98+99)-10000}\end{aligned}$

Ungrouping the terms,

$\begin{aligned}\longrightarrow\ \ &\sf{1^2+2^2-3^2-4^2+5^2+6^2-7^2-8^2+\,\dots\,+98^2-99^2-100^2}\\\\=\ \ &\sf{1-2-3+4+5-6-7+8+\,\dots\,+97-98-99-10000}\end{aligned}$

And grouping as the following:

$\begin{aligned}\longrightarrow\ \ &\sf{1^2+2^2-3^2-4^2+5^2+6^2-7^2-8^2+\,\dots\,+98^2-99^2-100^2}\\\\=\ \ &\sf{(1-2)-(3-4)+(5-6)-(7-8)+\,\dots\,+(97-98)-99-10000}\end{aligned}$

$\begin{aligned}\longrightarrow\ \ &\sf{1^2+2^2-3^2-4^2+5^2+6^2-7^2-8^2+\,\dots\,+98^2-99^2-100^2}\\\\=\ \ &\underbrace{\sf{(-1)-(-1)+(-1)-(-1)+\,\dots\,+(-1)}}_{\sf{49\ terms}}-\,\sf{99-10000}\end{aligned}$

$\begin{aligned}\longrightarrow\ \ &\sf{1^2+2^2-3^2-4^2+5^2+6^2-7^2-8^2+\,\dots\,+98^2-99^2-100^2}\\\\=\ \ &\underbrace{\sf{-1+1-1+1-\,\dots\,-1}}_{\sf{49\ terms}}-\,\sf{99-10000}\end{aligned}$

What if we separate the last -1 from the 49 terms?

$\begin{aligned}\longrightarrow\ \ &\sf{1^2+2^2-3^2-4^2+5^2+6^2-7^2-8^2+\,\dots\,+98^2-99^2-100^2}\\\\=\ \ &\underbrace{\sf{-1+1-1+1-\,\dots\,-1+1}}_{\sf{48\ terms}}-\,\sf{1-99-10000}\end{aligned}$

The entire 48 terms equal zero since $\sf{-1+1=0.}$ Therefore,

$\begin{aligned}\longrightarrow\ \ &\sf{1^2+2^2-3^2-4^2+5^2+6^2-7^2-8^2+\,\dots\,+98^2-99^2-100^2}\\\\=\ \ &\sf{-1-99-10000}\end{aligned}$

$\longrightarrow\sf{\underline{\underline{1^2+2^2-3^2-4^2+5^2+6^2-7^2-8^2+\,\dots\,+98^2-99^2-100^2=-10100}}}$

Hence $\bf{-10100}$ is the answer.