Question

In a series , if tn = n²-1/n+1 , then S6-S8=

Answer 1

Answer: -13

Step-by-step explanation:

tₙ = (n²-1)/(n+1)

∴, t₁ = (1²-1)/(1+1) = 0

   t₆ = (6²-1)/(6+1) = 35/7 = 5

   t₈ = (8²-1)/(8+1) = 63/9 = 7

Now, Sₙ = (n/2) x (t₁ + tₙ)

So, S₆ = (6/2) x (t₁ + t₆) = 3 x (0 + 5) = 15

     S₈ = (8/2) x (t₁ + t₈) = 4 x (0 + 7) = 28

∴, S₆ - S₈ = 15 - 28 = -13