#\lim _{n\to \infty }\sum _{i=1}^n\frac{3}{n}[(\frac{i}{n})^2+1]#.........??

2 Answers
MattyMatty
Feb 11, 2018

#4#

Explanation:

#= lim_{n->oo} (3/n^3) [ sum_{i=1}^{i=n} i^2 ] + (3/n) [ sum_{i=1}^{i=n} 1 ] #
#"(Faulhaber's formula)"#
#= lim_{n->oo} (3/n^3) [ (n(n+1)(2n+1))/6 ] + (3/n) [ n ] #
#= lim_{n->oo} (3/n^3) [ n^3/3 + n^2/2 + n/6 ] + (3/n) [ n ] #
#= lim_{n->oo} [1 + ((3/2))/n + ((1/2))/n^2 + 3]#
#= lim_{n->oo} [ 1 + 0 + 0 + 3 ]#
#= 4#

Ratnaker Mehta
Feb 11, 2018

# 4#.

Explanation:

Here is another way to solve the Problem :

Recall that, #int_0^1f(x)dx=lim_(n to oo)sum_(i=1)^n1/nf(i/n)...(star)#.

#:." The Reqd. Lim.="lim_(n to oo)sum_(i=1)^n3/n{(i/n)^2+1}#,

#=3[lim_(n to oo)sum_(i=1)^n1/n{(i/n)^2+1}]#,

#=3int_0^1{(x)^2+1}dx............[because, (star)]#,

#=3[x^3/3+x]_0^1#,

#=[x^3+3x]_0^1#,

#=1^3+3xx1-(0^3+3xx0)#,

#rArr " The Reqd. Lim.="4#.

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