Exercise 50 Page 415

Let's start by recalling the formula for this special series. Sum of squares of firstn positive integers ∑_(i=1)^n i^2 = n(n+1)(2n+1)/6Let's now consider the given series. ∑_(n=1)^(18) n^2 Here, we have that n = 18. Therefore, to find the desired sum, we will substitute 18 for n in the corresponding formula.

∑_(i=1)^n i^2 = n(n+1)(2n+1)/6

Substitute

n= 18

∑_(i=1)^(18) i^2 = 18( 18+1)(2( 18)+1)/18

▼

Evaluate right-hand side

Multiply

Multiply

∑_(i=1)^n i^2 = 18(18+1)(36+1)/6

AddTerms

Add terms

∑_(i=1)^n i^2 = 18(19)(37)/6

SplitIntoFactors

Split into factors

∑_(i=1)^n i^2 = 6 * 3(19)(37)/6

CancelCommonFac

Cancel out common factors

∑_(i=1)^n i^2 = 6 * 3(19)(37)/6

SimpQuot

Simplify quotient

∑_(i=1)^n i^2 = 3(19)(37)

Multiply

Multiply

∑_(i=1)^n i^2 = 2109