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1. In class we've talked about the method of differences. Here is a quick

summary. Suppose a₁, a2,... is a sequence of numbers and you want

to calculate 1 a. If you can miraculously find another sequence

b₁,b₂,... such that a; = b; - bi+1 then

a₂ = Σ(bi-bi+1) = (b₁-b₂)+(b₂-b3)++(bn-1-bn)+(bn-bn+1)

which is obviously equal to b₁-bn+1 because everything but the very

first and the very last terms cancel. So you get a hopefully nice formula

for the original sum:

n

n

Σ a=b₁-bn+1.

i=1

(a) Use the method of differences to find a formula for

(3i² + 3i+1).

i=1

Hint: 3² + 3i+ 1 = (i + 1)³ - i³.

(b) Given that i = n(n+1), use your answer to (a) to prove

that

Σ²²= n(n+1)(2n+1).

i=1

2 Letr be a real number with r +1

Fig: 1