$1^2 - 2^2 + 3^2 - 4^2 + \dots + 1999^2$ using closed form formula for sum of squares of first $n$ natural numbers

Solution 1:

If you first add all the squares: $$ 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + \cdots $$ and then subtract the even squares: $$ \phantom{1^2} - 2^2\phantom{ + 3^2} - 4^2 \phantom{+ 5^2} - 6^2 \phantom{-}\cdots $$ then you are left with only the odd squares: $$ 1^2 \phantom{+ 2^2} + 3^2 \phantom{+ 4^2} + 5^2 \phantom{+ 6^2} + \cdots $$ You need to subtract the even squares one more time to get $$ 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + \cdots $$

Solution 2:

We can use the identity $$(x+1)^2-x^2=x+1+x$$ to tackle the sum in 2 simpler ways:

Way 1 $$ \begin{aligned}&1^2 - 2^2 + 3^2 - 4^2 + 5^2-... -1998^2+ 1999^2 \\=& 1+(2+3)+(4+5)+…+(1998+1999) \\= &19990000\end{aligned}$$

Way 2 $$ \begin{aligned}&(1^2 - 2^2 + 3^2 - 4^2 + 5^2-6^2...-1998^2)+ 1999^2 \\=&-((1+2)+(3+4)+(5+6)+…+(1997+1998))+1999^2 \\= &19990000\end{aligned}$$

Solution 3:

Use the decomposition of squares into triangular numbers, $n^2=T_{n-1}+T_n$, and telescope: $$1^2-2^2+3^2-\cdots+1999^2=1-(1+3)+(3+6)-(6+10)+\cdots+(T_{1998}+T_{1999})$$ $$=T_{1999}=\frac{1999×2000}2=1999000$$