The question is simple, the solution, not so much
Q.Find the sum of the given expression $1^2- 2^2 + 3^2 – 4^2 + \dots + 1999^2$
My idea is we know $1^2 + 2^2 + 3^2 + 4^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$
So for $n=1999$ I get the sum as $2,66,46,67,000$
From this I need to subtract the squares of the even terms twice because subtracting once leaves with only the sum of the squares of the odd nos.
I observed something :
$2^2 + 4^2 + 6^2 + \dots + 1998^2 = (2 \cdot 1)^2 + (2 \cdot 2)^2 + \dots + (2 \cdot 999)^2$
Therefore to obtain the sum of the square of the even terms, I can take $4$ as common and use the aforementioned formula for $n=999$ and multiply it by $4$.
therefore sum of square of even terms = $1,33,13,34,000$
I need to subtract this sum twice to get the answer, because subtracting once simply leaves me with the sum of the squares of the odd numbers.
The answer is now $1999000$, which still doesn't match the answer key. Can someone explain where I am going wrong ?
Best Answer
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 $$ 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + \cdots $$