Question

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

Answer 1

Note: This problem is to be solved by using proof by induction!!

Answer 2

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.

Answer 3

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

Answer 4

the r.h.s of the equation is supposed to be 4^(n), NOT 4(n)--my mistake!

Answer 5

but what about the induction step now??..this is where i get stuck

Answer 6

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

Answer 7

how did you go from the 3rd to the 4th line and 4th to 5th and 5th to 6th and 6th to 7th???

Answer 8

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) ?

Answer 9

THANK YOU!! THANK YOU!!

Answer 10

But where did the 4^(k+1) in line 4?