2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8

2(2+2+2^2+2^3+2^4+2^5+2^6+2^7)
2^2(2+2+2^2+2^3+2^4+2^5+2^6)
2^3(2+2+2^2+2^3+2^4+2^5)
2^4(2+2+2^2+2^3+2^4)
2^5(2+2+2^2+2^3)
2^6(2+2+2^2)
2^7(2 + 2)
2^8(2)
2^9

That solution is actually not quite right.

There is an error in the first line of work shown (pulling a 2 out would not leave you with 2's on the inside like that).

Later on the 2^8 line: 2^7*(2+2) does not equal 2^8

I think the easiest way to do this is to evaluate the first handful, then factor out of the remainder and evaluate again.

2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8

2+2+4+8+16 + (2^4*(2+2^2+2^3+2^4))

32 + (16*(30))

32 + 480

512

2^9

Does that make sense?

Thank you so much. Now i understand it...

There are many ways to reach the solution, but I agree that grouping them and multiplying is the easiest one.

the first time when i saw this question in my test itself I was kind of taken by surprise...but in a moment i found this pattern:

2+2 = 2^2
2^2 + 2^2 = 2^3
2^3 + 2^3= 2^4

r u able to see a pattern, previous two terms equate to next term in series...

that way the last two will be : 2^8 + 2^8 = 2^9

milqman wrote:That solution is actually not quite right.

Later on the 2^8 line: 2^7*(2+2) does not equal 2^8

2^7 *(4) = 2^8 *(2)

I just showed it as 2^7 *(2+2)

Pulling out the 2 leaves you with 2^8 * (1+1) = 2^8 * (2)