If the sum of the first m terms of the series 1^3 1 + 1^3 + 3^3 7 + 1^3 + 3^3 + 5^3 17 + is 385 , then the value of m is equal to ______.

Step-by-step solution

Let the n^ th term of the given series be T_n . The numerator of T_n is the sum of the cubes of the first n odd natural numbers: Numerator = _ k=1 ^n (2k-1)^3 We know that (2k-1)^3 = 8k^3 - 12k^2 + 6k - 1 . Summing this from k=1 to n , we get: _ k=1 ^n (2k-1)^3 = 8 ( n(n+1) 2 )^2 - 12 ( n(n+1)(2n+1) 6 ) + 6 ( n(n+1) 2 ) - n = 2n^2(n+1)^2 - 2n(n+1)(2n+1) + 3n(n+1) - n = n [ 2n(n^2+2n+1) - 2(2n^2+3n+1) + 3n + 3 - 1 ] = n [ 2n^3 + 4n^2 + 2n - 4n^2 - 6n - 2 + 3n + 2 ] = n [ 2n^3 - n ] = n^2(2n^2 - 1) The denominator of