1.

`S=sum_(i=1)^nsum_(j=1)^isum_(k=1)^j1`

Answer» `sum_(i=1)^(n)sum_(j=1)^(i)sum_(k=1)^(j)1=sum_(i=1)^(n)(i(i+1))/(2)`
`=(1)/(2)[sum_(i=1)^(n)i^(2)+sum_(i=1)^(i)i]=(1)/(2)[sumn^(2)+sumn]`
`=(1)/(2)[(n(n+1)(2n+1))/(6)+(n(n+1))/(2)]`
`=(n(n+1))/(12)[2n+1+3]=(n(n+1)(n+2))/(6)`


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