# Rewriting (1 + 2 + + [k + 1])^2

This is from Spivak Calculus 4th ed.

Why are these two formulas equal?

Code:
(1 + 2 + ... + [k + 1])[SUP]2[/SUP] = (1 + 2 + ... + k)[SUP]2[/SUP] + 2(1 + ... + k)(k + 1) + (k + 1)[SUP]2[/SUP]
I'm especially confused about the right hand side. Where did 2(1 + ... + k)(k + 1) come from?

jedishrfu
Mentor
This is from Spivak Calculus 4th ed.

Why are these two formulas equal?

Code:
(1 + 2 + ... + [k + 1])[SUP]2[/SUP] = (1 + 2 + ... + k)[SUP]2[/SUP] + 2(1 + ... + k)(k + 1) + (k + 1)[SUP]2[/SUP]
I'm especially confused about the right hand side. Where did 2(1 + ... + k)(k + 1) come from?

Didn't they just group them into two terms?

(x + (k + 1))^2 = x^2 + 2*x*(k+1) + (k+1)^2

where x = 1 + 2 ... + k

LCKurtz
Homework Helper
Gold Member
This is from Spivak Calculus 4th ed.

Why are these two formulas equal?

Code:
(1 + 2 + ... + [k + 1])[SUP]2[/SUP] = (1 + 2 + ... + k)[SUP]2[/SUP] + 2(1 + ... + k)(k + 1) + (k + 1)[SUP]2[/SUP]
I'm especially confused about the right hand side. Where did 2(1 + ... + k)(k + 1) come from?

Try it with k = 3. Start with$$(1+2+3+4)(1+2+3+4) = [(1+2+3)+4)][(1+2+3)+4]$$Expand as a binomial without simplifying and see what happens. That might give you an idea.

 I didn't see the other post while I was typing.