Let T_k = 1 2 + 4 5 + 7 8 + + (3k-2)(3k-1) 3k^2 - 1 . Then the value of _ n ( 100 n^2 _ k=1 ^n T_k ) is equal to ______.

Step-by-step solution

First, we simplify the numerator of the general term T_k . The numerator is S = _ r=1 ^k (3r-2)(3r-1) . Expanding the terms inside the summation: (3r-2)(3r-1) = 9r^2 - 9r + 2 Now, summing from r=1 to k : S = 9 _ r=1 ^k r^2 - 9 _ r=1 ^k r + _ r=1 ^k 2 S = 9 ( k(k+1)(2k+1) 6 ) - 9 ( k(k+1) 2 ) + 2k S = 3 2 k(k+1)(2k+1) - 9 2 k(k+1) + 2k Taking k 2 common: S = k 2 [ 3(2k^2 + 3k + 1) - 9(k+1) + 4 ] S = k 2 [ 6k^2 + 9k + 3 - 9k - 9 + 4 ] S = k 2 [ 6k^2 - 2 ] = k(3k^2 - 1) The denominator of T_k is given as 3k^2 - 1 . Th