If Σᵣ₌₁¹⁰ r!(r³ + 6r² + 2r + 5) = α (11!), then the value of α is equal to ___________.

Solution

Maths sir · 5 steps

Find a quadratic g so each term is (r+1)! g(r+1) - r! g(r), and the whole sum collapses.

1. Aim to telescope. Look for f(r) = r! g(r) with f(r+1) - f(r) equal to the general term. Since (r+1)! = (r+1) r!, this difference is r![(r+1)g(r+1) - g(r)]. So you need (r+1)g(r+1) - g(r) = r³ + 6r² + 2r + 5.
2. Try a quadratic. The right side is cubic, so take g(r) = ar² + br + c. Expanding (r+1)g(r+1) - g(r) and collecting powers of r gives:

   (r+1)g(r+1) - g(r) = ar³ + (2a+b)r² + (3a+b+c)r + (a+b)

3. Match coefficients. a = 1, then 2 + b = 6 gives b = 4, then 3 + 4 + c = 2 gives c = -5. The constant term checks out: a + b = 5. So g(r) = r² + 4r - 5 = (r-1)(r+5).
4. Collapse the sum. All middle terms cancel, leaving 11! g(11) - 1! g(1). Here g(1) = 0 and g(11) = 121 + 44 - 5 = 160.
5. So. The sum is 160 × 11!, so α = 160.

The idea to keep. For a sum of r! times a polynomial, guess r! g(r) with g one degree lower. Use (r+1)g(r+1) - g(r) to match coefficients, then only the end terms survive.

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