Proove: n (-1) ^ r (n choose r) = 0 r= 0 & n (n choose r) = 2 ^ n r= 0 (Hint: Co
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Proove: n (-1) ^ r (n choose r) = 0 r= 0 & n (n choose r) = 2 ^ n r= 0 (Hint: Consider (1-1)^n & (1+1) ^ n) <---- given inproblem I am not sure what they are looking for? Thanks for yourhelp! Proove: n (-1) ^ r (n choose r) = 0 r= 0 & (-1) ^ r (n choose r) = 0 r= 0 & n (n choose r) = 2 ^ n r= 0 (Hint: Consider (1-1)^n & (1+1) ^ n) <---- given inproblem I am not sure what they are looking for? Thanks for yourhelp! (n choose r) = 2 ^ n r= 0 (Hint: Consider (1-1)^n & (1+1) ^ n) <---- given inproblem I am not sure what they are looking for? Thanks for yourhelp!Explanation / Answer
Let us consider the series... (1 + x)^n = sum( C(n, k) * x^k, k = 1..n) If we let x = -1, this simplifies to.. sum( C(n, k) * x^k, k = 1..n) sum( C(n, k) * (-1)^k, k = 1..n) (1 + (-1))^n 0^n 0 Thus, we have proven that the alternating sum of binomialcoefficients is 0. If we let x = 1, this simplifies to... sum( C(n, k) * x^k, k = 1..n) sum( C(n, k) * 1^k, k = 1..n) sum( C(n, k), k = 1..n) (1 + 1)^n 2^n Thus, we have proven that the sum of binomial coefficients is2^n.
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