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#