If 1² ·( ¹⁵ C₁)+2² ·( ¹⁵ C₂)+3² ·( ¹⁵ C₃)+…+15² ·( ¹⁵ C₁₅)=2ᵐ · 3ⁿ · 5ᵏ, where m, n, k ∈ N, then m+n+k is equal to :

Solution

Maths sir · 4 steps

Split r² as r(r-1) + r, and each piece turns into a clean power of 2.

1. Break up r squared. Write r² = r(r-1) + r. Then the sum splits into Σ r(r-1) ⁿCᵣ plus Σ r ⁿCᵣ. Both of these are standard results.
2. Use the two identities. Since r ⁿCᵣ = n ⁿ⁻¹Cᵣ₋₁, you get Σ r ⁿCᵣ = n 2ⁿ⁻¹. Applying it twice gives Σ r(r-1) ⁿCᵣ = n(n-1) 2ⁿ⁻².

   Σᵣ₌₁ⁿ r² ⁿCᵣ = n(n-1) 2ⁿ⁻² + n 2ⁿ⁻¹ = n(n+1) 2ⁿ⁻²

3. Put n = 15. The sum is 15 × 16 × 2¹³ = 15 × 2⁴ × 2¹³ = 2¹⁷ × 3 × 5.
4. So. m = 17, n = 1, k = 1, so m + n + k = 19. The answer is (b).

The idea to keep. When you see r² ⁿCᵣ, write r² = r(r-1) + r. Each factor of r pulls out an n and lowers the power of 2 by one, so the sum is n(n+1) 2ⁿ⁻².

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