• February 27th 2010, 10:04 AM
OsbourneOz
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!
• February 27th 2010, 10:22 AM
chisigma
The sum of the first n terms of a 'geometric progression' is...

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

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

Kind regards

$\chi$ $\sigma$
• February 27th 2010, 11:33 AM
OsbourneOz
Sorry, but I am totally lost!

Oz
• February 27th 2010, 11:50 AM
chisigma
Taking into account the identity...

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

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

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

... so that...

Kind regards

$\chi$ $\sigma$
• February 27th 2010, 01:39 PM
Laurent
Quote:

Originally Posted by OsbourneOz
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

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