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Question: find summation r=1 to r=39 (40C_r+1) (-1)^r...

find summation r=1 to r=39 (40C_r+1) (-1)^r

Answer

-39

Explanation

Solution

The problem asks us to find the sum ∑r=139(40Cr+1)(−1)r\sum_{r=1}^{39} (40C_{r+1}) (-1)^r.

Let's expand the sum: For r=1r=1: (40C1+1)(−1)1=−40C2(40C_{1+1}) (-1)^1 = -40C_2 For r=2r=2: (40C2+1)(−1)2=+40C3(40C_{2+1}) (-1)^2 = +40C_3 For r=3r=3: (40C3+1)(−1)3=−40C4(40C_{3+1}) (-1)^3 = -40C_4 ... For r=39r=39: (40C39+1)(−1)39=−40C40(40C_{39+1}) (-1)^{39} = -40C_{40}

So, the sum SS is: S=−40C2+40C3−40C4+⋯−40C40S = -40C_2 + 40C_3 - 40C_4 + \dots - 40C_{40}.

We know the binomial expansion of (1−x)n(1-x)^n. For n=40n=40 and x=1x=1: (1−1)40=∑k=04040Ck(−1)k(1)40−k(1-1)^{40} = \sum_{k=0}^{40} 40C_k (-1)^k (1)^{40-k} 040=40C0−40C1+40C2−40C3+40C4−⋯+(−1)3940C39+(−1)4040C400^{40} = 40C_0 - 40C_1 + 40C_2 - 40C_3 + 40C_4 - \dots + (-1)^{39} 40C_{39} + (-1)^{40} 40C_{40} 0=40C0−40C1+40C2−40C3+40C4−⋯−40C39+40C400 = 40C_0 - 40C_1 + 40C_2 - 40C_3 + 40C_4 - \dots - 40C_{39} + 40C_{40}.

Let's rearrange the terms in this identity to isolate the sum SS. 0=40C0−40C1+(40C2−40C3+40C4−⋯+40C40)0 = 40C_0 - 40C_1 + (40C_2 - 40C_3 + 40C_4 - \dots + 40C_{40}).

The terms in the parenthesis are −(−40C2+40C3−40C4+⋯−40C40)-( -40C_2 + 40C_3 - 40C_4 + \dots - 40C_{40} ). This means the terms in the parenthesis are −S-S.

So, the identity becomes: 0=40C0−40C1−S0 = 40C_0 - 40C_1 - S.

Now, we can solve for SS: S=40C0−40C1S = 40C_0 - 40C_1.

We know that nC0=1nC_0 = 1 and nC1=nnC_1 = n. So, 40C0=140C_0 = 1 and 40C1=4040C_1 = 40.

Substitute these values into the equation for SS: S=1−40S = 1 - 40 S=−39S = -39.

The final answer is −39\boxed{-39}.

Solution: The given sum is S=∑r=139(40Cr+1)(−1)rS = \sum_{r=1}^{39} (40C_{r+1}) (-1)^r. Let k=r+1k = r+1. When r=1,k=2r=1, k=2. When r=39,k=40r=39, k=40. The sum can be rewritten as: S=∑k=240(40Ck)(−1)k−1S = \sum_{k=2}^{40} (40C_k) (-1)^{k-1} S=−40C2+40C3−40C4+⋯−40C40S = -40C_2 + 40C_3 - 40C_4 + \dots - 40C_{40}.

We know the binomial identity: ∑k=0n(−1)knCk=nC0−nC1+nC2−⋯+(−1)nnCn=0\sum_{k=0}^{n} (-1)^k nC_k = nC_0 - nC_1 + nC_2 - \dots + (-1)^n nC_n = 0 (for n≥1n \ge 1). For n=40n=40: 40C0−40C1+40C2−40C3+40C4−⋯−40C39+40C40=040C_0 - 40C_1 + 40C_2 - 40C_3 + 40C_4 - \dots - 40C_{39} + 40C_{40} = 0.

Rearrange the terms to relate to SS: 40C0−40C1+(40C2−40C3+40C4−⋯+40C40)=040C_0 - 40C_1 + (40C_2 - 40C_3 + 40C_4 - \dots + 40C_{40}) = 0. The expression in the parenthesis is −(−40C2+40C3−40C4+⋯−40C40)-( -40C_2 + 40C_3 - 40C_4 + \dots - 40C_{40} ), which is −S-S. So, 40C0−40C1−S=040C_0 - 40C_1 - S = 0. S=40C0−40C1S = 40C_0 - 40C_1. S=1−40S = 1 - 40. S=−39S = -39.

The final answer is −39\boxed{-39}.