# Math Help - Summation

1. ## Summation

How do you calculate the summation when the top is n-1?

For example, if I want to find the general term of the sequence $\{ a_n \}1,3,7,13,21,31,...$

The first order difference is $\{ b_n \} 2,4,6,8,10,...$

$\{ b_n \}=2n$

$\{a_n \}=a_1+ \sum_{k=1}^{n-1} b_k=\sum_{k=1}^{n-1} 2k$

How do I calculate the sum of k?

2. Each term = n^2 - n + 1 ; if n=6, term=31

3. ${a_n}=1+\sum_{k=1}^{n-1}2k=1+2\sum_{k=1}^{n-1}k=1+2\frac{(n-1)n}{2}=1+(n-1)(n)$

4. Same as mine, Krahl; n=6, 1 + 6(5) = 31 ; not a sum

5. So for $\sum_{k=1}^{n-1} k=\frac {n(n-1)}{2}$? How do you calculate that?

How about $k^2$?

6. What do you mean?
The sum of the first n natural numbers is

$1+2+3+...+n=\frac{(n)(n+1)}{2}$

so the sum of the first n-1 natural numbers is

$1+2+3+...+(n-1)=(n-1)(n-1+1)=\frac{(n-1)(n)}{2}$

and the sum of the first n squared numbers is $\frac{n(n+1)(2n+1)}{6}$

But these two are different formulae

7. Originally Posted by Krahl
What do you mean?
The sum of the first n natural numbers is

1+2+3+...+n=(n)(n+1)

so the sum of the first n-1 natural numbers is

1+2+3+...+(n-1)=(n-1)(n-1+1)=(n-1)(n)

and the sum of the first n squared numbers is $\frac{n(n+1)(2n+1)}{6}$

But these two are different formulae
The formula you stated for $k^2$was for $\sum_{k=1}^n$, not n-1

EDIT: so if you want to find the sum for n-1, you just substitute n-1 as n into the regular formulas?

8. yes so just replace n in that formula with n-1.Thats all you need to do. try it out see if it works

so it becomes $\frac{(n-1)(n-1+1)(2(n-1)+1)}{6}$

is that ok?

so it becomes $\frac{(n-1)(n-1+1)(2(n-1)+1)}{6}$