If _ x 0 ( _ k=1 ^n (kx) ) e^ 2x^2 - (2x) = 35 2 , where n is a positive integer, then the value of n is

Step-by-step solution

The expression in the numerator can be rewritten as: ( _ k=1 ^n (kx) ) = _ k=1 ^n ( (kx)) For small u , the Maclaurin series expansion gives ( u) u^2 2 . Thus, the numerator is approximately: _ k=1 ^n (kx)^2 2 = x^2 2 _ k=1 ^n k^2 = x^2 2 n(n+1)(2n+1) 6 = x^2 n(n+1)(2n+1) 12 For the denominator, using the series expansions e^u 1 + u and u 1 - u^2 2 : e^ 2x^2 1 + 2x^2 (2x) 1 - (2x)^2 2 = 1 - 2x^2 So, e^ 2x^2 - (2x) (1 + 2x^2) - (1 - 2x^2) = 4x^2 Evaluating the limit: _ x 0 x^2 n(n+1)(2n+1) 12 4x^2 = n(n+1)(2n+1) 48