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sum(k=m)^n kCr...

`sum_(k=m)^n kC_r`

A

`""^(n+1)C_(r+1)`

B

`""^(n+1)C_(r+1)-""^(m)C_(r)`

C

`""^(n+1)C_(r+1)-""^(m)C_(r+1)`

D

`""^(n+1)C_(r+1)+""^(m)C_(r+1)`

Text Solution

Verified by Experts

For any integer `iger`, we have
`""^(r)C_(r)+""^(r+1)C_(r)+""^(1+2)C_(r)+.........+""^(i)C_(r)`
`""^(r+1)C_(r+1)+""^(r+1)C_(r)+.........+""^(i)C_(r)" "[:'""^(r)C_(r)=""^(r+1)C_(r+1)]`
`""^(r+2)C_(r+1)+""^(r+2)C_(r)+.........+""^(i)C_(r)`
`""^(r+3)C_(r+1)+""^(r+3)C_(r)+.........+""^(i)C_(r)`
`'=.......=""^(i+1)C_(r+1)`
Thus, we have
`sum_(k=r)^(i)""^(k)C_(r)=""^(i+1)C_(r+1)" "....(i)`
`:.sum_(k=m)^(n)""^(k)C_(r)=sum_(k=r)^(n)""^(k)C_(r)-sum_(k=r)^(m-1)""^(k)C_(r)`
`impliessum_(n=m)^(n)""^(k)C_(r)=""^(n+1)C_(r+1)-""^(m-1)C_(r+1)" "["Using (i)']`
`impliessum_(k=m)^(n)""^(k)C_(r)=""^(n+1)C_(r+1)-""^(m)C_(r+1)`
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OBJECTIVE RD SHARMA ENGLISH-PERMUTATIONS AND COMBINATIONS-Chapter Test
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