Let α = ∑ k = 0 n C k n 2 k + 1 and β = ∑ k = 0 n - 1 C k n C k + 1 n k + 2 . If 5 α = 6 β , then n equals

Step-by-step solution

Given, α = ∑ k = 0 n C k n 2 k + 1 ⇒ α = ∑ k = 0 n C k n · C k n k + 1 · n + 1 n + 1 Now, using C r + 1 n + 1 = n + 1 r + 1 · C r n we get, ⇒ α = 1 n + 1 ∑ k = 0 n C k + 1 n + 1 · C n - k n Now, using ∑ k = 0 n C r + 1 n + 1 · C n - r n = C n + 1 2 n + 1 we get, ⇒ α = 1 n + 1 · C n + 1 2 n + 1 . . . . . . . . i Now, solving β = ∑ k = 0 n - 1 C k n · C k + 1 n k + 2 n + 1 n + 1 ⇒ β = 1 n + 1 ∑ k = 0 n - 1 C n - k n · C k + 2 n + 1 Now, using ∑ k = 0 n C r + 2 n + 1 · C n - r n = C n + 2 2 n + 1 ⇒ β = 1 n + 1 · C n +