
Originally Posted by
sterps
I need some help with mathematical induction. It's driving me insane.
Use mathematical induction to prove that:
$\displaystyle 1/3 + 2/3^2 + 3/3^3 +...+ n/3^n = 1/4 (3 - 2n + 3/ 3^n )$
Basis step: Prove 1 exists in S.
n = 1 LHS =$\displaystyle 1/3^1 = 1/3$
RHS = $\displaystyle 1/4 (3 - 2(1) + 3/ 3^1 )$ = 1/3
So 1 exists in S
Inductive step:
Assume k exists in S. We must prove k+1 exists in S
So assume $\displaystyle 1/3 + 2/3^2 + 3/3^3 +...+ k/3^k = 1/4 (3 - 2k + 3/ 3^k )$
LHS = $\displaystyle 1/3 + 2/3^2 + 3/3^3 +...+ k/3^k + (k + 1 /3^K+1)$
RHS = $\displaystyle 1/4 (3 - 2k + 3/ 3^k ) + (k + 1 /3^K+1)$
After this i just get lost, i think i have a problem when trying to perform arithmetic on it. Any help would be much appreciated.
I also have another question:
Find a fourth order polynomial with integer coefficients for which $\displaystyle 1 + 2root3 - root7$ is a root.
SO we have $\displaystyle x = 1 + 2root3 - root7$
$\displaystyle x - 1 =2root3 - root7$
$\displaystyle (x - 1)^2 =12 - 7 + root21$
$\displaystyle (x^2 - 2x - 3)^2 = 21$
$\displaystyle (x^4 - 2x^2 +12 = 0$
any help would be much appreciated!