Σᵣ₌₁²⁰(r²+1)(r !) is equal to

Solution

Maths sir · 4 steps

Split r²+1 as r(r+1)-(r-1), and each term becomes a difference of two neighbours.

1. Split the coefficient. Write r²+1 = r(r+1)-(r-1). Multiply by r! and use (r+1) r! = (r+1)!.

   (r²+1) r! = r (r+1)! - (r-1) r!

2. Spot the telescope. Let g(r) = (r-1) r!. Then g(r+1) = r (r+1)!, so each term is g(r+1)-g(r). In the sum, everything cancels except g(21)-g(1).
3. Find the ends. g(21) = 20 × 21! and g(1) = 0 × 1! = 0. So the sum is 20 × 21!.
4. So. 20 × 21! = (22-2) × 21! = 22! - 2(21!), which is option (b).

The idea to keep. For a polynomial times r!, rewrite the polynomial so that the term becomes g(r+1)-g(r) with g(r) = (something) r!. Check it on r = 1 before you trust it.

Answer from the NTA final answer key. Solution by PrepWiser.