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sum(i=1)^n sum(j=1)^t sum(k=1)^j is equa...

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

A

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

B

`[(n(n+1))/2]^2`

C

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

D

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

Text Solution

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The correct Answer is:
D
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Knowledge Check

  • The value of sum_(i=1)^(n) sum_(j=1)^(i) sum_(k=1)^(j) 1 is

    A
    `sumn`
    B
    `sumn^(2)`
    C
    `sumn^(3)`
    D
    none of these

  • sum_(i = 1)^(n) sum_(j = 1)^(i) sum_(k = 1)^(j) is equa to

    A
    `(n (n + 1))/(2)`
    B
    `((n (n + 1))/(2))^(2)`
    C
    `(n (n + 1) (2n + 1))/(6)`
    D
    `(n (n + 1) (n + 2))/(6)`

  • sum_(i=1)^(n) sum_(i=1)^(n) i is equal to

    A
    `(n(n+1))/(2)`
    B
    `(n(n+1)^(2))/2`
    C
    `(n^(2)(n+1))/2`
    D
    none of these

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