A sequence is defined by a₁ = 4 and a_ n+1 = a_n^2 + a_n for n 1 . If S = _ n=1 ^ 1 a_n+1 , then the value of 100 S is equal to

Step-by-step solution

The given recurrence relation is a_ n+1 = a_n^2 + a_n . We can factor the right side as: a_ n+1 = a_n(a_n+1) Dividing both sides by a_n a_ n+1 , we get: 1 a_n(a_n+1) = 1 a_ n+1 Using partial fractions on the left side: 1 a_n - 1 a_n+1 = 1 a_ n+1 Rearranging this equation to isolate the term in our sum: 1 a_n+1 = 1 a_n - 1 a_ n+1 Now, consider the partial sum S_N = _ n=1 ^ N 1 a_n+1 : S_N = ( 1 a₁ - 1 a₂ ) + ( 1 a₂ - 1 a₃ ) + + ( 1 a_N - 1 a_ N+1 ) This is a telescoping sum. All intermediate terms cancel out: S_N =