I need to prove by induction on n, that

4^1 + 4^2 + 4^3 +....+ 4^n = 4/3 (4^n) - 1

For all integers 'n' is = or > than 1

Thank you very much!

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- Feb 27th 2010, 09:04 AMOsbourneOz'Proof' help needed please...
I need to prove by induction on n, that

4^1 + 4^2 + 4^3 +....+ 4^n = 4/3 (4^n) - 1

For all integers 'n' is = or > than 1

Thank you very much! - Feb 27th 2010, 09:22 AMchisigma
The sum of the first n terms of a 'geometric progression' is...

$\displaystyle S_{n} = a_{0}\cdot (1 + r + r^{2} + ... r^{n-1}) = a_{0}\cdot \frac{r^{n}-1}{r-1}$ (1)

Setting in (1) $\displaystyle a_{0}= r = 4$ what do You obtain?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Feb 27th 2010, 10:33 AMOsbourneOz
Sorry, but I am totally lost!

Oz - Feb 27th 2010, 10:50 AMchisigma
Taking into account the identity...

$\displaystyle r^{n}-1 = (r-1)\cdot (1 + r + r^{2} + \dots + r^{n-1})$ (1)

... you arrive to demostrate that is...

$\displaystyle S_{n} = a_{0}\cdot (1 + r + r^{2} + \dots + r^{n-1})= a_{0}\cdot \frac {r^{n}-1}{r-1}$ (2)

... so that...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Feb 27th 2010, 12:39 PMLaurent
In a proof by induction, you have to

a) check the initial case: here, you have to check the identity when $\displaystyle n=1$.

b) perform the induction step: assume that, for some $\displaystyle n\geq 1$, $\displaystyle 4^1+\cdots+4^n=\frac{4}{3}4^n-1$; then you have to prove that $\displaystyle 4^1+4^2+\cdots+4^{n+1}=\frac{4}{3}4^{n+1}-1$ (same identity with $\displaystyle n+1$ instead of $\displaystyle n$). To do this, simply note that $\displaystyle 4^1+\cdots+4^{n+1}=(4^1+\cdots+4^n)+4^{n+1}=(\frac {4}{3}4^n-1)+4^{n+1}$ (using the assumption), and after simplification you should get what you need. This will conclude the induction.