Question

help with a series lol

Answer 1

\[\sum_{n=1}^{\infty} 4^(n+1)/7^(n-1)\]
find the convergence... the parenthesis is to the power idk why it didn't show that way

Answer 2

you need { } around exponents

Answer 3

oh ok lol

Answer 4

\[\sum_{n=1}^{\infty}\frac{ 4^{n+1}}{7^{n-1}}\]

Answer 5

mhm

Answer 6

Try the geometric series test?

Answer 7

\[\sum_{n=1}^{\infty}\frac{ 4^{n+1}}{7^{n-1}} =4 (7) \sum_{i=1}^{\infty}(\frac{4}{7})^n\]
agree with batman

Answer 8

um why did you make it i =1 ?

Answer 9

you would rather it start at i=0 is what you are saying

Answer 10

you changed it from n=1 to i=1 lol ?

Answer 11

He means why did you go n = 1 to i = 1, no reason just a small mistake :P

Answer 12

oh that was a type-o

Answer 13

Oh okay I see what you did, thanks! =D

Answer 14

so is the answer (28) (4/3) <---answer after using the formula or did i do something wrong ?

Answer 15

you did a silly algebra mistake which lead me to the incorrect answer lol

Answer 16

huh? What did I do wrong?

Answer 17

\[\sum_{n=1}^{\infty}\frac{4^{n+1}}{7^{n-1}}=\sum_{n=1}^{\infty} \frac{4^n 4}{7^n7^{-1}}=4(7)\sum_{n=1}^{\infty} (\frac{4}{7})^n \]
there is nothing wrong here
I used law of exponents correctly

Answer 18

\[\sum_{n=1}^{\infty}\frac{4^{n+1}}{7^{n-1}}=\sum_{n=1}^{\infty} \frac{4^n 4}{7^n7^{-1}}=4(7)\sum_{n=1}^{\infty} (\frac{4}{7})^n\]
\[4(7)\sum_{n-1=0}^{\infty} (\frac{4}{7})^n \\ \text{ let } a=n-1 \\ 4(7)\sum_{a=0}^{\infty}(\frac{4}{7})^{a+1}=4(7) \frac{4}{7} \sum_{a=0}^{\infty}(\frac{4}{7})^a\]
maybe you didn't look at the right formula

Answer 19

You did it right, he probably gave you the wrong numbers.