Can you help me with this one? Take it slow please....
Show by induction: 2+2^2+2^3+.....+2^n=2*(2^n-1)
Thanks for all of your answers !
Hello, Neffets!
LaTeX isn't back yet . . . Hope you can read my notation . . .
Show by induction: .2 + 2^2 + 2^3 + . . . + 2^n .= .2(2^1 - 1}
Verify S(1): .2 .= .2(2^1 - 1) . . . yes!
Assume S(k): .2 + 2^2 + 2^3 + . . . + 2^k .= .2(2^k - 1)
Add 2^{k+1} to both sides:
. . 2 + 2^2 + 2^3 + . . . + 2^k + 2^{k+1} .= .2(2^k - 1) + 2^{k+1}
The right side is: .2^{k+1} - 2 + 2^{k+1} .= .2·2^{k+1} - 2 .= .2(2^{k+1} - 1)
And we have: .2 + 2^2 + 2^3 + . . . + 2^{k+1} .= .2(2^{k+1} - 1)
. . There!
Look at the previous line, on the right hand side of the "=" sign you
have:
2(2^k - 1) + 2^{k+1},
Now multiply the first bracket out to get:
(2^{k+1} - 2) + 2^{k+1}=2^{k+1} - 2 + 2^{k+1}
which is what you asked about, which Soroban then simplifies a bit
and puts back equal to the left hand side back at the earlier equation
where he had picked up the right hand side.
RonL