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Question
Answer the following:
If `(1 + 2 + 3 + 4 + 5 + ... "upto n terms")/(1 xx 2 + 2 xx3 + 3 xx 4 + 4 xx5 + ... "upto n terms") = 3/22` Find the value of n
Solution
`(1 + 2 + 3 + 4 + 5 + ... "upto n terms")/(1 xx 2 + 2 xx3 + 3 xx 4 + 4 xx5 + ... "upto n terms") = 3/22`
∴ `(sum_("r" = 1)^"n""r")/(sum_("r" = 1)^"n" "r"("r" + 1)) = 3/22`
∴ `(sum_("r" = 1)^"n" "r")/(sum_("r" = 1)^"n" ("r"^2 + "r")) = 3/22`
∴ `(sum_("r" = 1)^"n" "r")/(sum_("r" = 1)^"n" "r"^2 + sum_("r" = 1)^"n" "r") = 3/22`
∴ `([("n"("n" + 1))/2])/([("n"("n" + 1)(2"n" + 1))/6] + [("n"("n" + 1))/2]) = 3/22`
∴ `([("n"("n" + 1))/2])/("n"(("n" + 1))/2[(2"n" + 1)/3 + 1]) =3/22`
∴ `1/(((2"n" + 1 + 3)/3)) = 3/22`
∴ `3/(2"n" + 4) = 3/22`
∴ `1/("n" + 2) = 1/11`
∴ n + 2 = 11
∴ n = 9
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