Here's a question from high school mathematics.
If $ 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + \dots + 100^2 = x $, then
($1^2 + 3^2 + 5^2 + \dots + 99^2$) is equal to ?
Options were:
(a) $\frac{x}{2}-2525$
(b) $\frac{x}{2}+5050$
(c) $\frac{x}{2}-5050$
(d) $\frac{x}{2}+2525$
Is there a way to solve this question without using the formula for the summation of the squares of the first $n$ natural numbers.
I am definitely sure there's an easy way to do this without using the formula – $\frac{n(n+1)(2n+1)}{6}$ or those other summation formula for summing up the first $n$ odd/even natural numbers.
The correct answer is (a).
I was able to solve(wrong) it by using the formula for the sum of square of first n natural numbers that I found online.
Here's my solution using those formulas:
$$x = \frac{n(n+1)(2n+1)}{6} $$
$$x = {100 * 101 * 201 \over 6} = 338350$$
$${x\over 2} = {338350\over2}$$
$${x\over 2} = 169175$$
$$ e = 2^2 + 4^2 + 6^2 + 8^2 + 10^2 + \dots + 100^2$$
$$ e = {2n(n+1)(2n+1)\over3} $$
$$ e = {2*100(101)(201)\over3}$$
$$ e = 166650 $$
$$ {x\over2} – e = {x\over2} – e $$
$$ {x\over2} – e = 169175-166650 $$
$$ {x\over2} – e = 2525$$
$$ {x\over2} – 2525 = e$$
We know that sum of square of n natural numbers – sum of square of n natural even numbers = sum of square of n natural odd numbers.
$$ x – ({x\over2} – 2525) = 1^2 + 3^2 + 5^2 + 7^2 + 9^2 + \dots + 99^2 $$
$$ {x\over2} + 2525 = 1^2 + 3^2 + 5^2 + 7^2 + 9^2 + \dots + 99^2 $$
But the correct answer should be ${x\over2}-2525$
Best Answer
Let $A=\sum_{k=1}^{50} (2k-1)^2$, and $B=\sum_{k=1}^{50} (2k)^2$. Then, $A$ is the answer you want and $A+B=x$.
What is $B-A$? Well, We can group that in pairs to get $$B-A=(2^2-1^2)+\cdots (100^2-99^2)$$ which is $$(2+1)(2-1)+(4+3)(4-3)+\cdots +(100+99)(100-99)=$$ $$=3+7+11+\cdots +199=\frac {50}2\times (3+199) =5050$$
To conclude: adding the two equations, $A+B=x$ and $A-B=-5050$ we get $$2A=x-5050\implies A=\frac x2-2025$$