Let S_k denote the sum of the first k terms of a sequence a₁, a₂, a₃, . If _ k=1 ^n S_k = n(n+1)(n+2)(n+3) 24 for all integers n 1 , then the value of 11 _ k=1 ¹⁰ 1 a_k is ____________.

Step-by-step solution

Let T_n = _ k=1 ^n S_k = n(n+1)(n+2)(n+3) 24 . We can find S_n by using the difference S_n = T_n - T_ n-1 for n 2 (and S₁ = T₁ ): S_n = n(n+1)(n+2)(n+3) 24 - (n-1)n(n+1)(n+2) 24 S_n = n(n+1)(n+2) 24 [ (n+3) - (n-1) ] S_n = n(n+1)(n+2) 24 4 = n(n+1)(n+2) 6 Next, we find the n -th term of the sequence a_n using a_n = S_n - S_ n-1 for n 2 (and a₁ = S₁ ): a_n = n(n+1)(n+2) 6 - (n-1)n(n+1) 6 a_n = n(n+1) 6 [ (n+2) - (n-1) ] a_n = n(n+1) 6 3 = n(n+1) 2 We need to evaluate 11 _ k=1 ¹⁰ 1 a_k . Substitute a_k = k(k+1) 2 int