Hey guys I can't solve the second part of this following question, i.e. 'hence, ...' onwards.
Prove by induction that, for all positive integers n, [(end value=n) (summation notation, sigma) (starting value: r=1)] (r+1)2^r = n[2^(n+1)].
Hence, deduce that [(end value=2n) (summation notation, sigma) (starting value: r=n+1)] [(r+1)2^r + 2^(n+1)] = n[2^(2n+2)].
Sorry it's so messy but I don't have the programs for 'typing math'. Thanks in advance if anybody could help!
Hello, margaritas!
I've got the first part . . .
Prove by induction that, for all positive integers n:
. .
Verify that is true.
. . . . . true!
Assume that is true:
. .
Add to both sides:
. .
The left side is: . . . . the left side of
The right side is: .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . the right side of
The inductive proof is complete.