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Expert-verified Found in: Page 851 ### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508 # Partial Sums Find the first six partial sums ${S}_{1}, {S}_{2}, {S}_{3}, {S}_{4}, {S}_{5}, {S}_{6}$ of the sequence whose nth term is given. $1, 3, 5, 7,$….

The partial sums are:

$\begin{array}{l}\text{Firstpartialsum}={S}_{1}=1.\\ \text{Secondpartialsum}={S}_{2}=4.\\ \text{Thirdpartialsum}={S}_{3}=9.\\ \text{Fourthpartialsum}={S}_{4}=16.\\ \text{Fifthpartialsum}={S}_{5}=25.\\ \text{Sixth partialsum}={S}_{6}=36.\end{array}$

See the step by step solution

## Step 1.  Given

The given sequence is 1, 3, 5, 7…

## Step 2.  To determine

We have to find the first six partial sums.

## Step 3.  Calculation

The given sequence is 1, 3, 5, 7…

We see that the terms are increasing by 2.

So, the first 6 terms of these sequence are 1, 3, 5, 7, 9, 11.

Hence,

$\begin{array}{l}\text{Firstpartialsum}={S}_{1}=1.\\ \text{Secondpartialsum}={S}_{2}=1+3=4.\\ \text{Thirdpartialsum}={S}_{3}=1+3+5=9.\\ \text{Fourthpartialsum}={S}_{4}=1+3+5+7=16.\\ \text{Fifthpartialsum}={S}_{5}=1+3+5+7+9=25.\\ \text{Sixth partialsum}={S}_{6}=1+3+5+7+9+11=36.\end{array}$ ### Want to see more solutions like these? 