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Question
Find the value of n∑r=1(−1)r+1n−1Cr−1
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Find the value of n∑r=1(−1)r+1n−1Cr−1
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Solution
n∑r=1(−1)r+1n−1Cr−1 = n−1C0 - n−1C1 + n−1C2 - n−1C3..........................
Consider the expansion of (1+x)n−1,
(1+x)n−1 = n−1C0 + n−1C1x + n−1C2x2 + n−1C3 x3..........................
When x = -1, we get the required sum
⇒ (1−1)n−1 = n−1C0 - n−1C1 + n−1C2 - n−1C3.....................
= 0
n∑r=1(−1)r+1n−1Cr−1 = n−1C0 - n−1C1 + n−1C2 - n−1C3..........................
Consider the expansion of (1+x)n−1,
(1+x)n−1 = n−1C0 + n−1C1x + n−1C2x2 + n−1C3 x3..........................
When x = -1, we get the required sum
⇒ (1−1)n−1 = n−1C0 - n−1C1 + n−1C2 - n−1C3.....................
= 0
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