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Question
Find the sum up to the 17th term of the series `1^3/1 + (1^3 + 2^3)/(1 + 3) + (1^3 + 2^3 + 3^3)/(1 + 3 + 5) + ...`
Solution
t1 = `1^3/1`
t3 = `(1^3 + 2^3)/(1 + 3)`
t3 = `(1^3 + 2^3 + 3^3)/(1 + 3 + 5)`
∴ tn = `(1^3 + 2^3 + 3^3 + ... :n:^3)/(1 + 3 + 5 + ... "n terms")`
= `(sum"n"^3)/("n"^2) ((2"n" - 1 + 1)/2)^2`
= `("n"^2("n" + 1)^2)/(4("n"^2))`
= `("n" + 1)^2/4`
= `("n"^2 + 2"n" + 1)/4`
∴ Sn = `sum"t"_"n" = 1/4 sum"n"^2 + 2"n" + 1`
= `1/4 sum"n"^2 + sum2"n" + sum1`
= `1/4[("n"("n" + 1)(2"n" + 1))/6 + (2("n")("n" + 1))/2 + "n"]`
To find S17 put n = 17
S17 = `1/4[(17 xx 18 xx 35)/6 + (2(17)(18))/2 + 17]`
= `1/4[17 xx 105 + 17 xx 18 + 17]`
= `(17(105 + 18 + 1))/4`
= 17 × 31
= 527
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