Bsrhrki

I have been asked to calculate a general formula for $1 + b + 2b^2 + 3b^3 + ... + Nb^N$

I think that a part of this is a geometric sequence, and I have rewritten this as

$f(x) = 1 + sum_i=1^n icdot x^i$

(I couldn't figure out a way to make the first term be 1 with the progression so I just removed it from the series instead)

When I look on Wolfram Alpha it says that the partial sum formula for $ sum_i=1^n icdot x^i$ is:

$$sum_i=1^n icdot x^i = frac(nx-n-1)x^n+1+x(1-x)^2$$

On this question, an answer said that the general formula for the sum of a finite geometric series is:

$$sum_k=0^n-1x^k = frac1-x^n1-x$$

But if I substitute my ($icdot x^i$) into the formula mentioned above I don't get the same value as the one Wolfram gives me.

Two questions:

1. Is it correct to pull the first term out of the series so it becomes a geometric series, or is there another way?


2. How did Wolfram Alpha calculate that expression?

Here's what I did to get the formula for partial sums of this series:

It was too much to type in LaTeX, so just did it on paper. Hope you don't mind.

The extra factor of $i$ or $k$ or whatever you call it means we no longer have a geometric series, but we can get one by differentiating with respect to $x$. You'll want to check powers of $x$ carefully too.

Your presumptions are wrong. The first series is not a geometric series. and substituting $i*x^i$ into $sum_k=0^N x^k$ is not the same as $sum_i=0^Ni*x^i$.

Try to solve the problem using the principle of induction:

Let us assume the above relation holds for $n$: $sum_i=0^n ix^i=frac(nx-n-1)x^n+1+x(1-x)^2equiv=A_n$.

Now let us show it must also holds for $n+1$ and therefore for all $ninmathcalN$.
Thus,

beginalign
sum_i=0^n+1ix^i&=underbracesum_i=0^nix^i_=A_n+(n+1)x^n+1\
&=A_n+(n+1)x^n+1\
&=frac(nx-n-1)x^n+1+x(1-x)^2+(n+1)x^n+1\
&=frac(nx-n-1)x^n+1+x(1-x)^2+(n+1)x^n+1frac(1-x)^2(1-x)^2\
&=frac(nx-n-1)x^n+1+x+(n+1)x^n+1*(x-1)^2(1-x)^2\
&=frac1(1-x)^2biggl((nx-n-1)x^n+1+x+(n+1)x^n+1(1+x^2-2x)biggr)\
&=frac1(1-x)^2biggl(x^n+2((n+1)x-n-2)+xbiggr)\
&=frac1(1-x)^2biggl(x^(n+1)+1((n+1)x-(n+1)-1)+xbiggr)
endalign

So if it holds for any $n$ it must hold for $n+1$ too and therefore for all natural numbers. Now, you only have to check that it holds for $n=0$ or $n=1$ and you are done.

Hint: Think derivatives w.r.t. $b,$:

$$1 + b + 2b^2 + 3b^3 + dotsm + Nb^N=bsum_k=1^Nk,b^k-1=bBigl(sum_k=0^Nb^kBigr)^!!'$$

Start with $$S=1 + b + 2b^2 + 3b^3 + cdots + Nb^N$$

Multiply by $b$ and then subtract to give
$$bS=b + b^2 + 2b^3 + 3b^4 + cdots +(N-1)b^N + Nb^N+1$$
$$(b-1)S = -1 - b^2 - b^3 - b^4 - cdots -b^N + Nb^N+1$$

Do that again
$$b(b-1)S = -b - b^3 - b^4 - b^5 - cdots -b^N-b^N+1 - Nb^N+2$$
$$(b-1)^2S = 1-b +b^2 -(N+1)b^N+1 + Nb^N+2$$

Finally simplify
$$S = dfrac1-b +b^2 -(N+1)b^N+1 + Nb^N+2(b-1)^2.$$

As an example, with $b=2$ and $N=3$ the first line gives $S=1+2+8+24=35$ while the final line gives $S=frac1-2+4-64+961^2= 35$