BuyFind

Single Variable Calculus: Concepts...

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

Single Variable Calculus: Concepts...

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

Solutions

Chapter F, Problem 33E
To determine

To find: The value of the sum i=1n(i+1)(i+2).

Expert Solution

Answer to Problem 33E

The value of the sum i=1n(i+1)(i+2) is n3(n2+6n+11).

Explanation of Solution

Definition used:

If am,am+1,...,an are real numbers and m and n are integers such that mn, then i=mnai=am+am+1+am+2++an1+an.

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:

By the above definition, the sum i=1n(i+1)(i+2) expressed as follows.

i=1n(i+1)(i+2)=i=1n(i2+3i+2)=i=1ni2+3i=1ni+i=1n2=(n(n+1)(2n+1)6)+3(n(n+1)2)+2n=n(n+1)6[(2n+1)+9]+2n

On further simplification the value of the sum i=1n(i+1)(i+2) becomes,

i=1n(i+1)(i+2)=n(n+1)3[n+5]+2n=n3((n+1)(n+5)+6)=n3(n2+6n+11)

Thus value of the sum i=1n(i+1)(i+2) is n3(n2+6n+11).

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