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Question
Find 122 + 132 + 142 + 152 + … + 202.
Solution
122 + 132 + 142 + 152 + … + 202
= (12 + 22 + 32 + 42 + ... + 202) – (12 + 22 + 32 + 42 + ... + 112)
= \[\displaystyle\sum_{r=1}^{20} r^2 - \displaystyle\sum_{r=1}^{11} r^2\]
= `(20(20 + 1)(2 xx 20 + 1))/6 - (11(11 + 1)(2 xx 11 + 1))/6`
= `(20 xx 21 xx 41)/6 - (11 xx 12 xx 23)/6`
= 2870 – 506 = 2364.
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