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

[ksc311]

is there a shortcut to answering this question?

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

[blue_lotus]

Whenever there is a series , either it is a Arithmetic progression of Geometric progression.
G.P(Geometic progression) is when the series is increasion by a ratio.

This problem is a G.P starting from the Second element as the first element is repeated.
Sum of terms in a G.P = [a(r^n - 1)] / (r-1) for r>1

Let me explain using this example:
= 2+ 2 + (2^2) ....
= 2 + G.P of a series ------------------------------(1)

Let us use the formula from above
a = first term of the series = 2
r = common ratio of the series = the multiplying factor of the series = 2
i.e you notice that the series is 2,4,8,16 ... each of the previous term is getting multiplied by 2
n = number of tern in the series = 8

Now using the formula = G.P = [ 2(2^8 -1)] /(2-1) = 2(2^8 - 1) = 2^9 - 2

Substitute G.P in equation 1
= 2+ G.P
=2+ (2^9 -2)
= 2^9


The solution looks long, because I tried to explain in detail.
Only by knowing the formula you can solve it in 20 sec.

[Guest]

Whenever there is a series , either it is a Arithmetic progression of Geometric progression.
G.P(Geometic progression) is when the series is increasion by a ratio.

This problem is a G.P starting from the Second element as the first element is repeated.
Sum of terms in a G.P = [a(r^n - 1)] / (r-1) for r>1

Let me explain using this example:
= 2+ 2 + (2^2) ....
= 2 + G.P of a series ------------------------------(1)

Let us use the formula from above
a = first term of the series = 2
r = common ratio of the series = the multiplying factor of the series = 2
i.e you notice that the series is 2,4,8,16 ... each of the previous term is getting multiplied by 2
n = number of tern in the series = 8

Now using the formula = G.P = [ 2(2^8 -1)] /(2-1) = 2(2^8 - 1) = 2^9 - 2

Substitute G.P in equation 1
= 2+ G.P
=2+ (2^9 -2)
= 2^9


The solution looks long, because I tried to explain in detail.
Only by knowing the formula you can solve it in 20 sec.

Wow - thats really good to know this formula otherwise it could take minutes to solve this.

BLUE LOTUS: Can you also please explain Arithmetic progression formula and what all the variables stand for? An example would really help! :lol:

Much APPRECIATED!

[blue_lotus]

Hi,

Let us take a problem for Arithmetic Progression

Q) Find the sum of 2 + 4 + 6 + 8 + 10

We can see that the terms are increasing by 2 from the previous term

Ans) Formula for finding the sum = Sn = [n(2a+(n-1)d)] /2
a = first term in the series = 2
d = the difference between any two terms = 2
n = number of terms = 5

Substitute in the formula = [ 5(2*2 + (5-1)*2)]/2
= [5(4 + 8)]/2
= 30


If the question above is to find the 7th term of the series
We can use the formula = a + (n-1 )d
= 2 + ( 7 -1 )*2
= 2 + 12
= 14


Cheers,

[StaceyKoprince]

Nice explanation!