The sum of the first n terms of a sequence t_n is given by S_n = n^4 + 6n^3 + 11n^2 + 6n 4 . If P_n = _ k=1 ^n 1 t_k , then the value of P₁₀ is

Step-by-step solution

Given the sum of the first n terms, S_n = n^4 + 6n^3 + 11n^2 + 6n 4 . We can factor the numerator: n^4 + 6n^3 + 11n^2 + 6n = n(n^3 + 6n^2 + 11n + 6) = n(n+1)(n+2)(n+3) Thus, S_n = n(n+1)(n+2)(n+3) 4 . The general term t_n is given by t_n = S_n - S_ n-1 for n 2 . t_n = n(n+1)(n+2)(n+3) 4 - (n-1)n(n+1)(n+2) 4 t_n = n(n+1)(n+2) 4 [ (n+3) - (n-1) ] = n(n+1)(n+2) 4 4 = n(n+1)(n+2) Now, we need to evaluate P₁₀ = _ k=1 ¹⁰ 1 t_k = _ k=1 ¹⁰ 1 k(k+1)(k+2) . We can rewrite the general term of this sum using partial fractions