# Stuck and Frustrated!!!!

• Feb 3rd 2009, 08:29 AM
bearej50
Stuck and Frustrated!!!!
Conjecture a formula for the sum 5 + 9 + 13 + ... + (4n + 1)
• Feb 3rd 2009, 09:19 AM
Soroban
Hello, bearej50!

If you know some summation formulas, it's easy.

. . $\sum ^n_{k=1} 1 \:=\:n\qquad\qquad\sum^n_{k=1}k \:=\:\frac{n(n+1)}{2}$

Quote:

Conjecture a formula for the sum: . $5 + 9 + 13 + \hdots + (4n + 1)$

We have: . $\sum^n_{k=1}(4k+1) \;=\;4\sum^n_{k=1}k + \sum^n_{k+1}1 \;=\;4\left(\frac{n(n+1)}{2}\right) + n \;=\;2n^2+3n$