Question

\sum _{n=1}^∞\left(7-b\right)=2
cal 3 anyone?

Answer 1

\[\sum_{n=1}^{\infty}(7-b)^n=2\]

Answer 2

I gotta find b

Answer 3

Idk how to solve but I'm commenting to find out how later on

Answer 4

such series converges if and only if the subsequent condition holds:
\[\left| {7 - b} \right| < 1\; \Rightarrow 6 < b < 8\]
and the value of the corresponding sum is:
\[\frac{1}{{1 - \left( {7 - b} \right)}}\]
so, we have to solve this equation, for \(b\):
\[\frac{1}{{1 - \left( {7 - b} \right)}} = 2\]

Answer 5

The key to this is knowing what this sum is equal to:

\[S=\sum_{n=1}^\infty x^n\]

and then plugging in your value of (7-b)=x into that. It'll hopefully become clear soon what I mean.

So let's write that sum above, S without hiding the terms:

\[S= x+x^2+x^3+\cdots\]
notice, if we multiply the whole thing by x we get:

\[xS=x^2+x^3+x^4+\cdots\]
Now if we add x to it, what do we get? We get the same sum!

\[xS+x= x+x^2+x^3+\cdots\]\[xS+x=S\]

So with a few steps of algebra of solving for S we get an alternate way of writing S,

\[S = \frac{x}{1-x}=\sum_{n=1}^\infty x^n\]

Now you can plug in stuff and solve hopefully.

Answer 6

oops.. I have made an error I'm very sorry @Ratedover Here is the right equation:
\[\Large \frac{{7 - b}}{{1 - \left( {7 - b} \right)}} = 2\]
thanks!! for your reply :) @Kainui

Answer 7

Heh yeah that pesky index starting at 0 instead of 1, I figured I'd just go ahead and explain it cause of that :P

Answer 8

yes! well done !! :) @Kainui