Question

Answer the following:

If  `(1 xx 3 + 2 xx 5 + 3 xx 7 + ...  "upto n terms")/(1^3 + 2^3 + 3^3 + ...  "upto n terms") = 5/9`, find the value of n

Answer 1

Consider the series 1·3 + 2·5 + 3·7 + ... upto n terms.

Each term of this series is a product of two numbers. The first numbers in the products are 1, 2, 3, ... 

Hence, the first number in the r^th product is r.

The second numbers in the products are 3, 5, 7, ... which are in A.P. with a = 3 and d = 2. Hence, the second number in the r^th product is

a + (r – 1)d = 3 +(r – 1)2 = 2r + 1

∴ the r^th term = t_r = r(2r + 1)

∴ 1·3 + 2·5 + 3·7 + ... to n terms

∴ `sum_("r" = 1)^"n" "t"_"r" = sum_("r" = 1)^"n" "r"(2"r" + 1)`

= `sum_("r" = 1)^"n" (2"r"^2 + "r")`

Now, `(1 xx 3 + 2 xx 5 + 3 xx 7 + ...  "upto n terms")/(1^3 + 2^3 + 3^3 + ...  "upto n terms") = 5/9`

∴ `(sum_("r" = 1)^"n" (2"r"^2 + "r"))/(sum_("r" = 1)^"n" "r"^3) = 5/9`

∴ `(2 sum_("r" = 1)^"n" "r"^2 + sum_("r" = 1)^"n" "r")/(sum_("r" = 1)^"n" "r"^3) = 5/9`

∴ `(2[("n"("n" + 1)(2"n" + 1))/6] + [("n"("n" + 1))/2])/([("n"^2("n" + 1)^2)/4]) = 5/9`

∴ `(("n"("n" + 1))/2[(2(2"n" + 1))/3 + 1])/([("n"^2("n" + 1)^2)/4]) =5/9`

∴ `([(4"n" + 2 + 3)/3])/([("n"("n" + 1))/2]) = 5/9`

∴ `(4"n" + 5)/3 xx 2/("n"^2 + "n") = 5/9`

∴ `(8"n" + 10)/(3"n"^2 + 3"n") = 5/9`

∴ 72n + 90 = 15n² + 15n

∴ 15n² – 57n – 90 = 0

∴ 15n² – 75n + 18n – 90 = 0

∴ 15n(n – 5) + 18(n – 5) = 0

∴ (n – 5)(15n + 18) = 0

∴ n – 5 = 0 or 15n + 18 = 0

∴ n = 5 or n = `(-18)/15`

But n ∈ N, 

∴ n ≠ `(-18)/15`

Hence, n = 5