# Thread: Induction of series P2

1. ## Induction of series P2

Could someone please help me with these questions? I just need to prove that n=K+1 using the assumption n=k

1) 1/(x-1) - 1/x - 1/x^2 - 1/x^3 -...-1/x^n = 1/[x^n(x-1)] for n=1,2,3...
2) 1+2x2+3x2^2 +...+nx2^(n-1) = 1 + (n-1)2^n for n=1,2,3...
3) 1.2+2^2.2^2+3^2.2^3+...+n^2.2^n = (n^2 -2n+3)2^(n+1) - 6
for n =1,2,3...

Thankyou ! =)

2. Hello Christina,

it helps if you use the Sum notation:

$
\sum_{k = 1}^n k = 1 + 2 + 3 + \cdots + n
$

The notation says that you add from k = 1 to n k. Anyway, it makes keeping track of the types of sums you're interested in easier. Example:

$
\sum_{k = 0}^n \frac{-1}{x^n} = \frac{1}{x^n(x - 1)}
$

I'm not sure if that's the relation you're looking at, but in general, these kinds of induction work as follows. You chek the induction for a base case, such as n = 0;

$
\sum_{k = 0}^{(0)} \frac{-1}{x^n} = \frac{1}{x^{(0)}(x - 1)}$

$\frac{-1}{x^{(0)}} = \frac{1}{x^{(0)}(x - 1)}$
$\frac{-1}{1} = \frac{1}{(x - 1)}$
$-1 \neq \frac{1}{(x - 1)}
$

That would actually be the end of the inductive proof unless you can find another base case, where the relationship is true. (Perhaps I didn't understand your question)

Then you accept as the hypothesis the relation and try and show that it's true for (n + 1) term. With sum notation it usually looks like this (I changed the starting point to k = 1):

$
\sum_{k = 1}^n \frac{-1}{x^n} = \frac{1}{x^n(x - 1)}$

$\sum_{k = 1}^{n = 1} \frac{-1}{x^n} \stackrel{?}{=} \frac{1}{x^{n + 1}(x - 1)}
$

You change the sum of n + 1 terms to:

$
\sum_{k = 1}^{n = 1} \frac{-1}{x^{n}} = (\sum_{k = 1}^n \frac{-1}{x^n}) + \frac{-1}{x^{n + 1}}
$

substitue the hypothesis for the sum of the first n terms and show that if the assumption is true for n, it is also true for (n + 1):
$
\frac{1}{x^n(x - 1)} + \frac{-1}{x^{n + 1}} = \frac{1}{x^{n + 1}(x - 1)}
$