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Question
Evaluate `sum_(k=1)^11 (2+3^k )`
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Solution
`sum_("k" = 1)^11 (2 + 3^"k") = (2 + 3) + (2 + 3^2) + (2 + 3^3) + ......`up to 11 terms
= `2 × 11 + (3 + 3^2 + 3^3 + ......` up to 11 terms)
= `22 + (3(3^11 - 1))/(3 - 1)` ......... `[∵ "a" = 3, "r" = 3, "S" = ("a"("r"^"n" - 1))/("r" - 1)]`
= `22 + 3/2 (3^11 - 1)`
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