Question

Need help with sigma notation

Answer 1

114,126 (nearest whole number)

Explanation:

Geometric sequence

General form of a geometric sequence:
`a_n=ar^(n-1)`

(where a is the first term and r is the common ratio)

Given:

`\displaystyle \sum^(29)_(n=2)80(1.23)^(n-2)`

The sigma notation means to find the sum of the given geometric series where the first term is when n = 2 and the last term is when n = 29.

First term (a)

Determine the first term by substituting n = 2 into the given expression:

`n=2 \implies 80(1.23)^(2-2)=80`

Common ratio (r)

From inspection, the common ratio is 1.23.

nth term

As the first term is when n = 2 and the last term is when n = 29, there is a total of 28 terms.

Therefore:

• a = 80
• r = 1.23
• n = 28

Sum of the first n terms of a geometric series:

`S_n=(a(1-r^n))/(1-r)`

Substituting the given values into the formula:

`\implies S_(28)=(80\left(1-1.23^(28)\right))/(1-1.23)=114125.7556`

Therefore, the sum of the given geometric series is 114,126 (nearest whole number)

Answer 2

• 114126

Explanation:

The given expression describes the GP with

The first term:

• t = 80(1.23²⁻²) = 80(1.23⁰) = 80

The common ratio:

• r = 1.23

It is required to find the sum of the terms from 2 (the bottom number of sigma) through 29 (the top number of sigma)

The number of terms is

• n = 29 - 2 + 1 = 28

The sum of the first 28 terms is

• Sₙ = t(rⁿ - 1)/(r - 1)
• S₂₈ = 80(1.23²⁸ - 1)/(1.23 - 1) = 114125.75573 ≈ 114126 (rounded)