Find the value of limit of {1+2+3+......to (3n+2)terms}/(n+1)^2 as n approaches to infinity?

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Answer 1 · 2017-12-12T13:47:03

The answer is $=9/2$

We need

$sum_(k=1)^n k=n/2(n+1)$

Therefore,

$sum_(k=1)^(3n+2) k=(3n+2)/2(1+3n+2)=(3n+2)/2(3n+3)$

$=(3/2)(n+1)(3n+2)$

So,

$(sum_(k=1)^(3n+2)k)/(n+1)^2= ((3/2)(n+1)(3n+2))/(n+1)^2$

$=3/2(3n+2)/(n+1)$

$=(9n+6)/(2n+2)$

Finally,

$lim_(n->oo)(sum_(k=1)^(3n+2)k)/(n+1)^2=lim_(n->oo)(9n+6)/(2n+2)$

$=lim_(n->oo)(9+6/n)/(2+2/n)$

$=9/2$