4^3+4^4+4^5+....+4^(n)=4(4(n)-16) all over three
*Note: the entire right hand side of the equals sign is over three
Note: This problem is to be solved by using proof by induction!!
Step #1
test to see if it is true for n = 3
LS = 4^3 = 64
RS = 4(4(3)-16)/3 = -16/3 ≠ LS
check the typing of your right side, if it is correctly typed your statement is false.
for the right hand side when i plugged 3 into it i got: 4(4*4^(3)-16)/3 which equals 4(64-16)/3 which equals 4*48/3, which is 64 also
the r.h.s of the equation is supposed to be 4^(n), NOT 4(n)--my mistake!
but what about the induction step now??..this is where i get stuck
That's better
so RS = 4[ 4^n - 16]/3
step2
assume it is true for n=k
that is,
4^3 + 4^4 + ... + 4^k = 4[ 4^k - 16]/3
step 3
prove that is it then true for n = k+1, that is
prove:
4^3 + 4^4 + .. + 4^k + 4^(k+1) = 4[ 4^(k+1) - 16]/3
LS = (4^3 + 4^4 + .. + 4^k) + 4^(k+1)
= 4[ 4^k - 16]/3 + 4^(k+1) , getting a common denominator ...
= [ 4(4^k - 16) + 3(4^(k+1))]/3
= [ 4(4^k - 16) + 3(4)(4^k) ]/3
= 4 [4^k - 16 + 3(4^k) ] /3
= 4 [ 4(4^k) - 16 ]/3
= 4 [4^(k+1) - 16]/3
= RS
YEAHH
how did you go from the 3rd to the 4th line and 4th to 5th and 5th to 6th and 6th to 7th???
form the third to the 4th line I took out a common factor of 4
the 4th line simply shows that common factor out front
for the 5th line I added 4^k and 3(4^k) which is 4(4^k) just like x + 3x = 4x
in the next line, isn't 4(4^k) = 4^1 x 4^k = 4^(k+1) ?