Exercise 2 Page 455

Let's start by recalling the formula for this special series. Sum of squares of firstk positive integers ∑_(n=1)^k n^2 = k(k+1)(2k+1)/6Let's now consider the given series. ∑_(n=1)^(16) n^2 Here, we have that k = 16. Therefore, to find the desired sum, we will substitute 16 for k in the corresponding formula.

∑_(n=1)^k n^2 = k(k+1)(2k+1)/6

Substitute

k= 16

∑_(n=1)^(16) n^2 = 16( 16+1)(2( 16)+1)/6

▼

Simplify right-hand side

Multiply

Multiply

∑_(n=1)^(16) n^2 = 16(16+1)(32+1)/6

AddTerms

Add terms

∑_(n=1)^(16) n^2 = 16(17)(33)/6

Multiply

Multiply

∑_(n=1)^(16) n^2 = 8976/6

CalcQuot

Calculate quotient

∑_(n=1)^(16) n^2 = 1496