Exercise 48 Page 415

Let's start by recalling the formula for this special series. Sum of firstn positive integers ∑_(i=1)^n i = n(n+1)/2Let's now consider the given series. ∑_(n=1)^(16) n Here we have that n=16. Therefore, to find the desired sum we will substitute 16 for n in the corresponding formula.

∑_(i=1)^n i = n(n+1)/2

Substitute

n= 16

∑_(i=1)^(16) i = 16( 16+1)/2

▼

Evaluate right-hand side

AddTerms

Add terms

∑_(i=1)^(16) i = 16(17)/2

SplitIntoFactors

Split into factors

∑_(i=1)^(16) i = 2* 8(17)/2

CancelCommonFac

Cancel out common factors

∑_(i=1)^(16) i = 2* 8(17)/2

SimpQuot

Simplify quotient

∑_(i=1)^(16) i = 8(17)

Multiply

Multiply

∑_(i=1)^(16) i = 136