"easy" exponent problem, please help

[tomintampa]

Why, how, what am i missing here, the answer is A? Or do I have the wrong answer, i would guess E


1)2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8 =
a.2^9
b.2^10
c.2^16
d.2^35
e.2^37

[bkparikh]

Looks like you are adding the exponents. You can only add exponents during multiplication with the same base, example: 2^8 * 2^8 = 2^16
general: a^m * a^n = a^(m+n)

When dealing with 2's, you'll notice that:
2^1 + 2^1 = 2^2
2^2 + 2^2 = 2^3
...
2^n + 2^n = 2^(n+1)

Using this, you see that your equation simplifies as follows:
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8
2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8
2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8
2^4 + 2^4 + 2^5 + 2^6 + 2^7 +2^8
...
2^8 + 2^8 = 2^9

Hope that helps!

[tomintampa]

yes, thank you very much. does this pattern hold for other numbers?

[bkparikh]

the pattern that holds is a*a^n = a^(n+1)
So in 2's case 2*2^n = 2^n + 2^n = 2^(n+1)
For 3's, 3*3^n = 3^n + 3^n + 3^n = 3^(n+1)

Hope that helps!

[abhishek_mumbai]

that was a good trick.. thanks for sharing..

[tomintampa]

thank you...i wasnt even thinking progressions!

[clock60]

to my mind it is simple geometric series with a1=2 and 8 members of the series. formula for the sum is Sn=a1*(Q^n-1/(Q-1).
we have 8 members (do not count first 2) and Q=2
S8=2*(2^8-1/(2-1)=2*(2^8 -1)=2^9 -2. then destroy first 2 with our and the answer 2^9

[wanderwanderer]

Hi
another way to solve ;
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8 =
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7(1+2)=
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 . 3 =
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 (1+ 2. 3)=
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 . 7 =
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 (1 + 2 .7)=
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 . 15

Beyond this u need not compute becoz u will notice the numbers are 3 , 7 , 15 which 2n + 1

so simply substitutin will give you
2 + 2 + 2^2 .127 = 2 ^2 . 128 = 2^9

[Crack_Mat]

Originally Posted by bkparikh

Looks like you are adding the exponents. You can only add exponents during multiplication with the same base, example: 2^8 * 2^8 = 2^16
general: a^m * a^n = a^(m+n)

When dealing with 2's, you'll notice that:
2^1 + 2^1 = 2^2
2^2 + 2^2 = 2^3
...
2^n + 2^n = 2^(n+1)

Using this, you see that your equation simplifies as follows:
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8
2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8
2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8
2^4 + 2^4 + 2^5 + 2^6 + 2^7 +2^8
...
2^8 + 2^8 = 2^9

Hope that helps!

Nice approach!!