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Question

# Solve: limn→∞1+3+6+........n(n+1)/2n3

A
12
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B
13
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C
16
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D
18
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Solution

## The correct option is C 16limn→∞1+3+6+…n(n+1)2n3=limn→∞n∑m=1m(m+1)2n3=limn→∞n∑m=1m2+m2n3=12limn→∞n∑m=1m2+n∑m=1mn3It is known that:n∑m=1m=n(n+1)2n∑m=1m2=n(n+1)(2n+1)6Hence,12limn→∞n∑m=1m2+n∑m=1mn3=limn→∞1(2n3)[n(n+1)2+n(n+1)(2n+1)6]=limn→∞1(2n3)[n(n+1)2]+limn→∞1(2n3)[n(n+1)(2n+1)6]In case of limn→∞if the degree of the polynomial in the numerator < degree of the denominator, the limit is 0. If the degrees of the numerator and denominator are equal, the limit is the ratio of the leading terms.Hence,=limn→∞1(2n3)[n(n+1)2]+limn→∞1(2n3)[n(n+1)(2n+1)6]=0+22×6=16

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