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Question
Answer the following:
Find 122 + 132 + 142 + 152 + ... 202
Solution
122 + 132 + 142 + 152 + ... 202
= (12 + 22 + 32 + 42 + … + 202) – (12 + 22 + 32 + 42 + … + 112)
= `sum_("r" = 1)^20 "r"^2 - 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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