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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?
Verified Answer
The sum rCr + r+1Cr + r+2Cr + .....+ nCr (n > r) equalsa)nCr +1b)n ...
C(n, r) + c(n -1, r) + C(n - 2, r) + ... + C(r, r)
= r+1Cr+1 + r+1Cr + r+2Cr + .... + n-1Cr + nCr
= n+1Cr+1 (applying same rule again and again ) (∴ nCr + nCr-1 = n+1Cr)
= r+1Cr+1 + r+1Cr + r+2Cr + .... + n-1Cr + nCr
= n+1Cr+1 (applying same rule again and again ) (∴ nCr + nCr-1 = n+1Cr)
Most Upvoted Answer
The sum rCr + r+1Cr + r+2Cr + .....+ nCr (n > r) equalsa)nCr +1b)n ...
C(n, r) + c(n -1, r) + C(n - 2, r) + ... + C(r, r)
= r+1Cr+1 + r+1Cr + r+2Cr + .... + n-1Cr + nCr
= n+1Cr+1 (applying same rule again and again ) (∴ nCr + nCr-1 = n+1Cr)
= r+1Cr+1 + r+1Cr + r+2Cr + .... + n-1Cr + nCr
= n+1Cr+1 (applying same rule again and again ) (∴ nCr + nCr-1 = n+1Cr)
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