Question

Find an expression for the general term of the series. Give the starting value of the index.

((x-1)^3)-(((x-1)^5)/(2!))+(((x-1)^7)/(4!))-(((x-1)^9)/(6!))+....

Answer 1

@agent0smith

Answer 2

expression: ((x-1)^(n+2))/2n! ???

Answer 3

im not sure what it means by the starting value of the index:/

Answer 4

hmm... i think you're almost right

2n! won't work though, then the first term would be over 2.

(2n-2)! works for all denominators.

Answer 5

almost! lol

Answer 6

starting index... n=1? i think...

Answer 7

im goin for it! lol thank you!

Answer 8

yeah (2n+2)! works cos 0! is 1, then all the others work too

Answer 9

wait! you're missing a negative!

it alternates, so maybe throw (-1)^n in front of it!!

Answer 10

@megannicole51

Answer 11

No... (-1)^(n+1) cos the first is positive.

Answer 12

ooooh okay! thank you!

Answer 13

wait so add n+1 to the den. as well?

Answer 14

no, just throw it in front of the whole thing, that should fix it.

Answer 15

I'm sure it's possible to simplify, but your TA loves you, so meh

Answer 16

lol ;)

Answer 17

:D

Answer 18

wait... the exponent on this -(-1)^n ((x-1)^(n+2))/(2n-2)!

should be -(-1)^n ((x-1)^(2n+1))/(2n-2)!

n+2 doesn't work, 2n+1 does