Question

Induction. What is wrong with this proof?
For all n in [0,1,2,3..),
1(3^0) + 3(3^1) + 5 (3^2) + 7 (3^3) + ... + (2n+1) 3^n = n 3^(n+1).

Proof:
Suppose n is a natural number, and suppose that 1(3^0) + 3(3^1) + 5 (3^2) + 7 (3^3) + ... + (2n+1) 3^n = n 3^(n+1). Then
1(3^0) + 3(3^1) + 5 (3^2) + 7 (3^3) + ... + (2n+1) 3^n + (2n + 3) 3^(n+1)

= n 3^(n+1) + (2n + 3) 3^(n+1)
= (3n+3) 3^(n+1)
= (n+1) (3) 3^(n+1)
= (n+1) 3^(n+2)

Answer 1

I realize there is no base case, which turns out to be incorrect, but why did the math work ?

Answer 2

when n = 0,
1 = 0, (which is wrong)

Answer 3

that the math works out and the base case doesn't means that the formula is off by a constant value

Answer 4

the constant value is equal to the base case; adding 1 makes it work

Answer 5

1(3^0) + 3(3^1) + 5 (3^2) + 7 (3^3) + ... + (2n+1) 3^n = n 3^(n+1)+1

Answer 6

ahh i see