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gettinit
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gettinit
2^(22)+2^(23)+2^(24)+.......+2^(43)+2^(44)=

a) 2^22(2^(23)-1)

b) 2^22(2^(24)-1)

c) 2^22(2^(25)-1)

d) 2^23(2^(21)-1)

e) 2^23(2^(22)-1)

Please explain methodology.

\(2^2^2 + 2^2^3 .... 2^4^4 = 2^2^2 ( 1 + 2 + 4 + 8 .... 2^2^2)\)

The part within parentheses is a geometric progression with a = 1 and r = 2, where there are 23 terms. So if you apply the geometric sum formula for this you'll get:
\(a \frac{( 1-r^n^-^1)}{(1-r)} = \frac{1-2^2^4}{1-2} = (2^2^4 - 1)\)

Thus I think the final answer is B.
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gettinit
2^(22)+2^(23)+2^(24)+.......+2^(43)+2^(44)=

a) 2^22(2^(23)-1)

b) 2^22(2^(24)-1)

c) 2^22(2^(25)-1)

d) 2^23(2^(21)-1)

e) 2^23(2^(22)-1)

Please explain methodology.

\(2^2^2 + 2^2^3 .... 2^4^4 = 2^2^2 ( 1 + 2 + 4 + 8 .... 2^2^2)\)

The part within parentheses is a geometric progression with a = 1 and r = 2, where there are 23 terms. So if you apply the geometric sum formula for this you'll get:
\(a \frac{( 1-r^n^-^1)}{(1-r)} = \frac{1-2^2^4}{1-2} = (2^2^4 - 1)\)

Thus I think the final answer is B.


Dude is that geometric progression formula correct?

Cheers
J :)
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gettinit
2^(22)+2^(23)+2^(24)+.......+2^(43)+2^(44)=

a) 2^22(2^(23)-1)

b) 2^22(2^(24)-1)

c) 2^22(2^(25)-1)

d) 2^23(2^(21)-1)

e) 2^23(2^(22)-1)

Please explain methodology.

\(2^2^2 + 2^2^3 .... 2^4^4 = 2^2^2 ( 1 + 2 + 4 + 8 .... 2^2^2)\)

The part within parentheses is a geometric progression with a = 1 and r = 2, where there are 23 terms. So if you apply the geometric sum formula for this you'll get:
\(a \frac{( 1-r^n^-^1)}{(1-r)} = \frac{1-2^2^4}{1-2} = (2^2^4 - 1)\)

Thus I think the final answer is B.


Dude is that geometric progression formula correct?

Cheers
J :)

Sum of GP where first term is A, common ratio is R and total number of terms is n is given by
\(Sum = \frac{A(1 - R^n)}{(1 - R)}\)
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gettinit
2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) =

A. 2^22(2^(23)-1)

B. 2^22(2^(24)-1)

C. 2^22(2^(25)-1)

D. 2^23(2^(21)-1)

E. 2^23(2^(22)-1)


How about this approach -

A = 2^(22) + 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) ... 1
2A = 2^(23) + 2^(24) + ... + 2^(43) + 2^(44) + 2^45 ... 2

2 - 1 gives
A = 2^45 - 2^22 = 2^22 ( 2^23 - 1)
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arabella
\(2^{(22)} + 2^{(23)} + 2^{(24)} + ... + 2^{(43)} + 2^{(44)} =\)


A \(2^{22}(2^{(23)}-1)\)

B \(2^{22}(2^{(24)}-1)\)

C \(2^{22}(2^{(25)}-1)\)

D \(2^{23}(2^{(21)}-1)\)

E \(2^{23}(2^{(22)}-1)\)


a different way and a point to remember.....
\(2^1 + 2^2 + 2^3 .........+ 2^{n-1} + 2^n=2^{n+1}-2\)

so...
\(2^{(22)} + 2^{(23)} + 2^{(24)} + ... + 2^{(43)} + 2^{(44)} = (2^1+2^2+......2^{44})-(2^1+2^2+....+2^{21}\)
as per above formula
\((2^1+2^2+......2^{44})-(2^1+2^2+....+2^{21}=(2^{(44+1)}-2)-(2^{(21+1)}-2)=2^{(44+1)}-2-2^{(21+1)}+2=2^{(45)}-2^{(22)}=2^{22}(2^{23}-1)\)
A

you can check why \(2^1 + 2^2 + 2^3 .........+ 2^{n-1} + 2^n=2^{n+1}-2\) at
https://gmatclub.com/forum/2-252919.html
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arabella
\(2^{(22)} + 2^{(23)} + 2^{(24)} + ... + 2^{(43)} + 2^{(44)} =\)


A \(2^{22}(2^{(23)}-1)\)

B \(2^{22}(2^{(24)}-1)\)

C \(2^{22}(2^{(25)}-1)\)

D \(2^{23}(2^{(21)}-1)\)

E \(2^{23}(2^{(22)}-1)\)

Let’s begin by factoring out 2^22, which is common to each of the terms:

2^22 + 2^23 + … + 2^44 = 2^22(1 + 2 + 2^2 + … + 2^22)

To find an expression for 1 + 2 + 2^2 + … + 2^22, we will use the formula

a^n - 1 = (a - 1)(a^(n - 1) + a^(n - 2) + … + a + 1).

If we let a = 2 and n = 23, we obtain

2^23 - 1 = (2 - 1)(2^22 + 2^21 + … + 2 + 1).

Thus:

2^22 + 2^23 + … + 2^44 = 2^22(1 + 2 + 2^2 + … + 2^22) = 2^22(2^23 - 1)

Answer: A
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2(22)+2(23)+2(24)+...+2(43)+2(44)=2(22)+2(23)+2(24)+...+2(43)+2(44)=

This can be re written as : \(2^{21}\cdot\left(2+2^2+.............2^{23\ \ }\right)\)
The numbers inside the bracket can be considered to be X.

\(2+2^2+.........2^{23}\ =\ X\)

\(2\cdot\left(1+..........+\ 2^{22}\right)\ =\ X\)
\(\ 2\cdot\left(X\ -2^{23}+1\right)\ =\ X\)
We get \(2X\ -\ 2^{24}+2\ \ =\ X\)
X = \(2\cdot\left(2^{23}-1\right)\)
Multiplying this with the other term we get :
\(2^{21}\cdot2\left(2^{23}-1\right)\ =\ 2^{22}\left(2^{23}-1\right)\)
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I tried something like this:

2^21 + 2^22..........+ 2^44 = X and we need to find X.

Adding 2^21 on both sides, we get:

2^21 + 2^21 + 2^22 .... 2^44 = X + 2^21

We know that 2^21 + 2^21 = 2^22, this way the equation becomes,

2^45 = X + 2^21

2^45 - 2^21 = X

2^21 ( 2^23 - 1 ) = X

Hopefully, this is a quicker method?
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