If Δ r = r 2 r - 1 3 r - 2 n 2 n - 1 a 1 2 n n - 1 n - 1 2 1 2 n - 1 3 n + 4 , then the value of ∑ r = 1 n − 1 Δ r

Step-by-step solution

Δ r = r 2 r - 1 3 r - 2 n 2 n - 1 a 1 2 n n - 1   n - 1 2 1 2 n - 1 3 n + 4 Since, the second and the third rows are independent of r , hence the sum is applied to the first row only. ⇒ ∑ r = 1 n − 1 Δ r = ∑ r = 1 n − 1 r 2 ∑ r = 1 n − 1 r − ∑ r = 1 n − 1 1 3 ∑ r = 1 n − 1 r − 2 ∑ r = 1 n − 1 1 n 2 n − 1 a 1 2 n n − 1 n − 1 2 1 2 n − 1 3 n + 4 Using ∑ r = 1 n r = n n + 1 2 , we get