# The value of lim n → ∞ ∑ i = 1 n 3 n [ ( 1 + 3 i n ) 3 − 2 ( 1 + 3 i n ) ] .

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter F, Problem 46E
To determine

## To find: The value of limn→∞∑i=1n3n[(1+3in)3−2(1+3in)].

Expert Solution

The value of limni=1n3n[(1+3in)32(1+3in)] is 1954.

### Explanation of Solution

Theorem used:

Let c be a constant and n be a positive integer. Then,

i=1nc=nc, i=1ni=n(n+1)2 and i=1ni2=n(n+1)(2n+1)6.

Calculation:

Simplify the expression limni=1n3n[(1+3in)32(1+3in)] and obtain the limit as follows.

limni=1n3n[(1+3in)32(1+3in)]=limni=1n3n[1+9in+27i2n2+27i3n326in]=limni=1n[81i3n4+81i2n3+9in23n]=limn[81n4(n(n+1)2)2+81n3(n(n+1)(2n+1)2)+9n2(n(n+1)2)3nn]=limn[81n4(n4(1+1n)24)+81n3(n3(1+1n)(2+1n)2)+9n2(n2(1+1n)2)3]

On further simplification the value of the limit becomes,

limni=1n3n[(1+3in)32(1+3in)]=limn[814(1+1n)2+812(1+1n)(2+1n)+92(1+1n)3]=814(1+1)2+812(1+1)(2+1)+92(1+1)3=814(1)+812(1)(2)+92(1)3=814+81+923=1954

Thus, the value of limni=1n3n[(1+3in)32(1+3in)] is 1954.

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