# How much is this sum?

1. Jan 29, 2013

### tsuwal

the answer is 3^2048. How do I get there?

2. Jan 29, 2013

### AlephZero

Since you are given the answer, use that information!

You have to prove that
$(2+3)(2^2+3^2)\cdots(2^{2048} + 3^{2048}) + 2^{4096} - 3^{4096} = 0$
Now, think what you can do with $2^{4096} - 3^{4096}$ ...

3. Jan 30, 2013

### tsuwal

i don't know, what can i do :S?

4. Jan 30, 2013

### tsuwal

and the answer is not given, it's multiple choice

5. Jan 30, 2013

### mathman

a2 - b2 = (a-b)(a+b)

Start with a = 22048 and b = 32048
next repeat with a = 21024 and b = 31024
etc.
At the end you will have (2-3)(2+3). Just be careful with the sign.

Last edited: Jan 30, 2013
6. Jan 30, 2013

### tsuwal

but i got a plus sign not a minus sign...

7. Jan 31, 2013

### HallsofIvy

Staff Emeritus
No, its a minus sign:
$(2+3)(2^2+3^2)\cdots(2^{2048} + 3^{2048}) + 2^{4096} - 3^{4096} = 0$
AlephZero was referring to the last pair on the left.

8. Jan 31, 2013

### tsuwal

now i get it. thanks!