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sum(i = 1)^n sum(j = 1)^i sum(k = 1)^j 1...

`sum_(i = 1)^n sum_(j = 1)^i sum_(k = 1)^j 1` is equal to

A

`(n(n+1)(n+2))/6`

B

`sum n^2`

C

`^nC_3`

D

`^(n + 2)C_3`

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The correct Answer is:
A, D
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