Undetermined Coefficients (Part 1)

Algebra Level 4

${ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+\cdots +{ 99 }^{ 2 } = \, ?$

1 solution

Sahil Bansal
Mar 24, 2016

Required sum is: $1^2+3^2+5^2+...+99^2=(1^2+2^2+3^2+...+100^2)-(2^2+4^2+6^2+...+100^2)$

$=\sum_{n=1}^{100}n^2 - \sum_{m=1}^{50}(2m)^2$

$=\sum_{n=1}^{100}n^2 - 4\sum_{m=1}^{50}m^2$

$=\frac{100*101*201}{6}-4*\frac{50*51*101}{6}$

$[\because \sum n^2=\frac{n(n+1)(2n+1)}{6}]$

Hence, required sum is: $\boxed{{\color{#456461}{\textbf{166650}} }}$

Did the same way

I Gede Arya Raditya Parameswara - 4 years, 4 months ago

Where did you get that (100 * 101 * 201)/6-(4 * 50 * 51 * 101)/6?

Manu Mehta - 5 years, 2 months ago

Using: $\sum n^2=\frac{n(n+1)(2n+1)}{6}$

Sahil Bansal - 5 years, 2 months ago