Question

I have (40n^2)+(9n). I WANT a formula where I can find the sum of a series. Like.. n=1+n=2+n=3

Answer 1

There is no formula for the sum of a series. You have to add however many partial sums. If it is an infinite series it is even more complicated.

Answer 2

...what? i wish people worded their questions more carefully.

are you asking for the sum from n to s of 40n^2 + 9n?

then it's 20s(s+1)(2s+1)/3 + 9s(s + 1)/2

= s(s + 1)/6[40(2s + 1) + 27]

= s(s + 1)(80s + 67)/6

Answer 3

from 1 to s*, sorry

Answer 4

If I understood correctly, you want\[S \doteq \sum_{n = 1}^m \left(40n^2 + 9n\right)\](summing the first m terms of the formula).

Since this is just a sum, we can factor out constants:\[S = 40\sum_{n = 1}^m n^2 + 9 \sum_{n = 1}^m n.\]

The first sum is easier. If we want to sum the first m numbers, the easiest way to do it is adding the numbers the following way:

\begin{array}a & 1 & 2 & 3 & 4 & \dots & m-2 & m-1 & m \\ & m & m-1 & m-2 & m-3 & \dots & 3 & 2 & 1\end{array}
If you add downwards, each sum is m+1 and there are m sums, so the total is m(m+1). Since we added each number two times, we simply divide by two and get
\[\sum_{n = 1}^m n = \frac{m(m+1)}{2}.\]
For the second part, we can prove by induction that\[\sum_{n = 1}^m n^2 = \frac{1}{6}m(m+1)(2m+1).\]If m=1, the sum is 1 and \[\frac{1}{6}1(1+1)(2+1)=1.\]If this is true for m,\begin{eqnarray*}\sum_{n = 1}^{m+1}n^2 &=& \sum_{n = 1}^m n^2 + (m+1)^2 \\&=& \frac{1}{6}m(m+1)(2m+1)+(m+1)^2\\&=&(m+1)\left(\frac{1}{6}m(2m+1)+m+1\right)\\&=&(m+1)\left(\frac{1}{6}\left(2m^2+7m+6\right)\right)\\&=&\frac{1}{6}(m+1)(m+2)(2m+3)\end{eqnarray*}which completes the proof.
Using these formulae,\[\sum_{n = 1}^m\left(40n^2+9n\right) = \frac{1}{6}m(m+1)(80m+67).\]