1^2+3^2+5^2+7^2+9^2 upto 19^2 = how much.how to solve?

for that u shd know the formula for sum of squares of natural nos.
1^2+2^2+..n^2=n(n+1)(2n+1)/6
so v vll divide p= 1^2+2^+..+19^2.. as sum of squares of odd no. and even nos.
lets say m= 1^2+3^2+5^2...19^2
n=2^2+4^2+6^2...+18^2
can we say p=m+n?
now u know the ans for p by the formula.
now n=(1*2)^2+(2*2)^2+...(9*2)^2 isnt it?
I can now safely take 2^2 common. so now n=2^2(1^2+2^2+3^2+...+9^2) this also u can find out by formula.
so next just do p-n and u are done!

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