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Question
Find the sum 22 + 42 + 62 + 82 + ... upto n terms.
Solution
22 + 42 + 62 + 82 + ... upto n terms
= (2 x 1)2 + (2 x 2)2 + (2 x 3)2 + (2 + x 4)2 + ...
= \[\displaystyle\sum_{r=1}^{n}(2r^2)\]
= 4\[\displaystyle\sum_{r=1}^{n} r^2\]
= `(4."n"("n" + 1)(2"n" + 1))/6`
= `(2"n"("n" + 1)(2"n" + 1))/3`.
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