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# Find the sum of the following series : $1^{2} + 2^{2} + 3^{2} + 4^{2} + .... + 20^{2}$

$(a)\;1280\qquad(b)\;2870\qquad(c)\;2180\qquad(d)\;5740$

Toolbox:
• Sum of squares of n natural numbers is given by $\large\frac{n}{6} (n+1) (2n+1)$
Sum of squares of n natural numbers is given by $\large\frac{n}{6} (n+1) (2n+1)$
Here n = 20. Replacing n = 20, n+1 = 21 and 2n+1 = 41,
We get Sum of series S = $\large\frac{20 X 21 X 41}{6}$
S = 2870

edited Nov 19, 2013