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प्रश्न
If S1, S2 and S3 are the sums of first n natural numbers, their squares and their cubes respectively then show that - 9S22 = S3 (1 + 8 S1)
उत्तर
S1, S2 and S3 are the sums of first n natural numbers, their squares and their cubes respectively.
∴ S1 = `sum_("r" = 1)^"n" "r" = ("n"("n" + 1))/2`
S2 = `sum_("r" = 1)^"n" "r"^2 = ("n"("n" + 1)(2"n" + 1))/6`
S3 = `sum_("r" = 1)^"n" "r"^3 = ("n"^2("n" + 1)^2)/4`
= S3 (1+ 8S1) = `("n"^2("n" + 1)^2)/4[1 + 8*("n"("n" + 1))/2]`
= `("n"^2("n" + 1)^2)/4[1 + 4"n"("n" + 1)]`
= `("n"^2("n" + 1)^2)/4(4"n"^2 + 4"n" + 1)`
= `("n"^2("n" + 1)^2(2"n" + 1)^2)/4`
= `9*("n"^2("n" + 1)^2(2"n" + 1)^2)/36`
= `9*[("n"("n" + 1)(2"n" + 1))/6]^2` = 9S22
Hence, 9S22 = S3(1 + 8S1).
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