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प्रश्न
Answer the following:
Find `1^2/1 + (1^2 + 2^2)/2 + (1^2 + 2^2 + 3^2)/3 + ...` upto n terms
उत्तर
Here, tr = `(1^2 + 2^2 + 3^2 + ... + "r"^2)/"r"`
= `(sum_("i" = 1)^"r" "i"^2)/"r"`
= `("r"("r" + 1)(2"r" + 1))/(6"r")`
= `(("r" + 1)(2"r" + 1))/6`
= `(2"r"^2 + 3"r" + 1)/6`
= `1/3"r"^2 + 1/2"r" + 1/6`
∴ `1^2/1 + (1^2 + 2^2)/2 + (1^2 + 2^2 + 3^2)/3 + ...` upto n terms
= `sum_("r" = 1)^"n" "t"_"r"`
= `sum_("r" = 1)^"n" (1/3"r"^2 + 1/2"r" + 1/6)`
= `1/3 sum_("r" = 1)^"n" "r"^2 + 1/2 sum_("r" = 1)^"n" "r" + 1/6 sum_("r" = 1)^"n" 1`
= `1/3*("n"("n" + 1)(2"n" + 1))/6 + 1/2*("n"("n" + 1))/2 + 1/6"n"`
= `"n"/2[(("n" + 1)(2"n" + 1))/9 + ("n" + 1)/2 + 1/3]`
= `"n"/2[(2(2"n"^2 + 3"n" + 1) + 9("n" + 1) + 6)/18]`
= `"n"/36(4"n"^2 + 6"n" + 2 + 9"n" + 9 + 6)`
= `("n"(4"n"^2 + 15"n" + 17))/36`.
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