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The sum rCr + r+1Cr + r+2Cr + .... + nCr (n > r) equals
- a)nCr+1
- b)n+1Cr+1
- c)n+1Cr-1
- d)n+1Cr
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The sum rCr + r+1Cr + r+2Cr + .... + nCr (n > r)equals a)nCr+1b)n+1...
C(n, r) + c(n ‐1, r) + C(n ‐ 2, r) + . . . + C(r, r)
= n+1Cr+1 (applying same rule again and again)
(∵ nCr + nCr‐1 = n+1Cr)
= n+1Cr+1 (applying same rule again and again)
(∵ nCr + nCr‐1 = n+1Cr)
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The sum rCr + r+1Cr + r+2Cr + .... + nCr (n > r)equals a)nCr+1b)n+1...
The sum rCr + r+1Cr + r+2Cr + .... + nCr (n > r) can be simplified using the hockey-stick identity.
The hockey-stick identity states that the sum of the combinations from rCr to nCr is equal to (n+1)C(r+1).
Therefore, the given sum can be simplified to (n+1)C(r+1).
The hockey-stick identity states that the sum of the combinations from rCr to nCr is equal to (n+1)C(r+1).
Therefore, the given sum can be simplified to (n+1)C(r+1).
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