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Prove by induction that, for all $x≠1$, $1+2x+3x^2+...+nx^{n-1} = \dfrac {nx^{n+1}-(n+1)(x^n)+1}{(x-1)^2}$

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  • $\begingroup$ What have you tried? $\endgroup$ – Viktor Glombik Sep 16 at 15:36
  • $\begingroup$ Please see math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Angina Seng Sep 16 at 15:37
  • $\begingroup$ I did the base case and got to (nx^(n+1)-(n+1)(x^n)+(n+1)(x-1)^2+1)/(x-1)^2 $\endgroup$ – Saiera Shueib Sep 16 at 15:56
  • $\begingroup$ If the base case is $n = 1$ how could you have gotten (nx^(n+1)-(n+1)(x^n)+(n+1)(x-1)^2+1)/(x-1)^2 ? Wouldn't you have gotten $1 = \frac {1x^2-2x+1}{(x-1)^2}$? $\endgroup$ – fleablood Sep 16 at 16:27
  • $\begingroup$ Induction on summations is EASY once you realize the simple trick that if $\sum_{k=1}^n a_k = f(n)$ is what we need to prove then the induction step is simply noting that $\sum_{k=1}^{n} a_k + a_{n+1} = f(n) + a_{n+1}$, we just have to prove $f(n) + a_{n+1}= f(n+1)$ which is usually straightforward. .. in this case we just have to prov $\dfrac {nx^{n+1}-(n+1)(x^n)+1}{(x-1)^2}+(n+1)x^n = \dfrac {(n+1)x^{n+2}-(n+2)(x^{n+1})+1}{(x-1)^2}$ which is just algebraic manipulation. I'm not sure why induction on sum means proving $f(n)+a_n = f(n+1)$ is not more widely understood. $\endgroup$ – fleablood Sep 16 at 16:35
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So base case would be to prove that if $n=1$ then $1 = \frac {1*x^2 - 2*x^1+1}{(x-1)^2}$. So prove that $\frac {x^2 -2x + 1}{(x-1)^2} = 1$ if $x \ne 1$.

And the induction step would be to prove if

$1 + 2x+ ....+nx^{n-1} = \dfrac {nx^{n+1}-(n+1)(x^n)+1}{(x-1)^2}$ then.

$(1 + 2x+ ....+nx^{n-1}) + (n+1)x^{n} =$

$ \dfrac {nx^{n+1}-(n+1)(x^n)+1}{(x-1)^2} + (n+1)x^{n}\underbrace{ =}_{\text{prove this equality}} \dfrac {(n+1)x^{n+2}-(n+2)(x^{n+1})+1}{(x-1)^2}$

So just do that. Prove that if $x \ne 1$ then $ \dfrac {nx^{n+1}-(n+1)(x^n)+1}{(x-1)^2} + (n+1)x^{n}= \dfrac {(n+1)x^{n+2}-(n+2)(x^{n+1})+1}{(x-1)^2}$

........

$ \dfrac {nx^{n+1}-(n+1)(x^n)+1}{(x-1)^2} + (n+1)x^{n}=$

$\dfrac {[nx^{n+1}-(n+1)(x^n)+1] + [(n+1)x^{n}(x-1)^2]}{(x-1)^2}=$

$\dfrac {[nx^{n+1}-(n+1)(x^n)+1] + [(n+1)x^{n+2}-2(n+1)x^{n+1} +(n+1)x^{n}]}{(x-1)^2}=$

$\dfrac {(n+1)x^{n+2}+[n-2(n+1)]x^{n+1} +[-(n+1)+(n+1)]x^{n} + 1}{(x-1)^2}=$

$\dfrac {(n+1)x^{n+2}-(n+2)x^{n+1} + 1}{(x-1)^2}$

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For $n=1$, the LHS is equal to $1$, and the RHS is equal to $$\dfrac {x^{2}-2x+1}{(x-1)^2}=1$$

so there is equality. Let's suppose that for an integer $n \geq 1$, one has $$1+2x+3x^2+...+nx^{n-1}=\dfrac {nx^{n+1}-(n+1)(x^n)+1}{(x-1)^2}$$

Then you have $$1+2x+3x^2+...+nx^{n-1} + (n+1)x^n = \dfrac {nx^{n+1}-(n+1)(x^n)+1}{(x-1)^2} + (n+1)x^n $$ $$ =\dfrac {nx^{n+1}-(n+1)(x^n)+1+(n+1)x^n(x-1)^2}{(x-1)^2} $$ $$= \dfrac {nx^{n+1}-(n+1)(x^n)+1+(n+1) (x^{n+2}-2x^{n+1}+x^n)}{(x-1)^2}$$ $$=\dfrac {(n+1) x^{n+2}-(n+2)x^{n+1}+1}{(x-1)^2}$$

which is the equality at the rank $n+1$. Therefore, by the induction principle, the equality is true for all $n \geq 1$.

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