# To prove: The formula ∑ i = 1 n i 3 = [ n ( n + 1 ) 2 ] 2 .

### 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 39E
To determine

## To prove: The formula ∑i=1ni3=[n(n+1)2]2.

Expert Solution

### Explanation of Solution

Consider the sum i=1n[(1+i)4i4].

i=1n[(1+i)4i4]=(2414)+(3424)+(4434)++((n+1)4n4)=(n+1)414=n4+4n3+6n2+4n+11=n4+4n3+6n2+4n

Thus, i=1n[(1+i)3i3]=n4+4n3+6n2+4n (1)

Simplify the sum i=1n[(1+i)4i4] by another method.

i=1n[(1+i)4i4]=i=1n[(i4+4i3+6i2+4i+1)i4]=i=1n[(i4+4i3+6i2+4i+1)i4]=i=1n4i3+6i2+4i+1=4i=1ni3+6i=1ni2+4i=1ni+i=1n1

Further simplified as,

i=1n[(1+i)3i3]=4i=1ni3+6(n(n+1)(2n+1)6)+4(n(n+1)2)+n=4i=1ni3+n(n+1)(2n+1)+2n(n+1)+1=4i=1ni3+2n3+3n2+n+2n2+2n+n=4i=1ni3+2n3+5n2+4n

That is, i=1n[(1+i)4i4]=4i=1ni3+2n3+5n2+4n (2)

Compare the equation (2) with equation (1).

Thus, 4i=1ni3+2n3+5n2+4n=n4+4n3+6n2+4n.

Simplify the above equation and obtain the formula for i=1ni3.

4i=1ni3=n4+4n3+6n2+4n(2n3+5n2+4n)=n4+2n3+n2=n2(n2+2n+1)=n2(n+1)2

Therefore, i=1ni3=[n(n+1)2]2.

Hence the formula is proved.

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