Conjecture a formula for the sum 5 + 9 + 13 + ... + (4n + 1)
Hello, bearej50!
If you know some summation formulas, it's easy.
. . $\displaystyle \sum ^n_{k=1} 1 \:=\:n\qquad\qquad\sum^n_{k=1}k \:=\:\frac{n(n+1)}{2}$
Conjecture a formula for the sum: .$\displaystyle 5 + 9 + 13 + \hdots + (4n + 1)$
We have: .$\displaystyle \sum^n_{k=1}(4k+1) \;=\;4\sum^n_{k=1}k + \sum^n_{k+1}1 \;=\;4\left(\frac{n(n+1)}{2}\right) + n \;=\;2n^2+3n$