# general term

• Aug 21st 2009, 02:54 AM
thereddevils
general term
Find the general term , or nth term , u_n .

-4 , 8 , -14 , 22 , -32

i can see te pattern ..

4 = 4
8 = 4 + (2x2)
14 = 4 + (2x2) + (2x3)
22 = 4 + (2x2) + (2x3) + (2x4)
32 = 4 + (2x2) + (2x3) + (2x4) + (2x5)

But i cant do the induction ??

i am really weak at questions like this ..
• Aug 21st 2009, 03:37 AM
running-gag
Hi

You can find an easy relation between $\displaystyle |u_{n+1}|$ and $\displaystyle |u_{n}|$ that will let you find the general expression of $\displaystyle |u_{n}|$ with respect to n
• Aug 21st 2009, 04:43 AM
HallsofIvy
Quote:

Originally Posted by thereddevils
Find the general term , or nth term , u_n .

-4 , 8 , -14 , 22 , -32

i can see te pattern ..

4 = 4
8 = 4 + (2x2)
14 = 4 + (2x2) + (2x3)
22 = 4 + (2x2) + (2x3) + (2x4)
32 = 4 + (2x2) + (2x3) + (2x4) + (2x5)

But i cant do the induction ??

i am really weak at questions like this ..

Then $\displaystyle u_n= 4+ (2x2)+ (2x3)+ \cdot\cdot\cdot+ (2*n)$ which could also be written as $\displaystyle u_n= 4+ 2\sum_{i= 1}^n i$

Now it will help to know that $\displaystyle \sum_{i=0}^n i= \frac{n(n-1)}{2}$ so that $\displaystyle \sum_{i=1}^n i= \frac{n(n-1)}{2}- 1= \frac{n^2-n}{2}- \frac{2}{2}= \frac{n^2- n- 2}{2}$
• Sep 1st 2009, 07:06 AM
thereddevils
Quote:

Originally Posted by running-gag
Hi

You can find an easy relation between $\displaystyle |u_{n+1}|$ and $\displaystyle |u_{n}|$ that will let you find the general expression of $\displaystyle |u_{n}|$ with respect to n

Thanks . But i am not sure how to do this .. Can u show it to me >>
• Sep 1st 2009, 08:59 AM
running-gag
$\displaystyle |u_1| = 4$
$\displaystyle |u_2| = 8 = |u_1| + 4 = |u_1| + 2 \times 2$
$\displaystyle |u_3| = 14 = |u_2| + 6 = |u_1| + 2 \times 3$
$\displaystyle |u_4| = 22 = |u_3| + 8 = |u_3| + 2 \times 4$
...
$\displaystyle |u_k| = |u_{k-1}| + 2 k$

$\displaystyle |u_k| - |u_{k-1}| = 2 k$

Sum for k = 2 to n
Spoiler:

$\displaystyle |u_n| - |u_1| = 2 \sum_{k=2}^{n} k$

$\displaystyle |u_n| - 4 = 2 \left(\frac{n(n+1)}{2}-1\right) = n^2 + n - 2$

$\displaystyle |u_n| = n^2 + n + 2$

And since the sign of $\displaystyle u_n$ is alternatively - and +

$\displaystyle u_n = (-1)^{n}(n^2 + n + 2)$
• Sep 2nd 2009, 07:11 AM