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Question

Find the value of expression ni=1ij=1jk=16.


A

n(n+1)(n+3)

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B

n(n+1)(n+2)

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C

n(n+2)(n+3)

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D

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Solution

The correct option is B

n(n+1)(n+2)


ni=1ij=1jk=16.

Let's calculate the summation part one by one from right hand side

= ni=1ij=16j. {Where jk=16=6j}

= 6ni=1 i(i+1)2 {Where jk=1 6j = 6 × sum of first n natural number}

= 3[ni=1i2+ni=1i] where ni=1i2 = n(n+1)(2n+1)6

= 3[n(n+1)(2n+1)6 + n(n+1)2]

= 32 n(n+1) [2n+13 + 1]

= 32 n(n+1) 2n+43 = n(n+1)(n+2)


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