I solved this this way,...
for odd values of k we get
1/2,1/8,1/32,....
for even values of k we get
-1/4,-1/16.-1/64,...
So adding the terms in their corresponding positions
we get
1/4 + 1/16+ 1/64
so its like a Geometric series..i then apply the GP formula ..then calculate (1/4)^5 ans so forth to get the actual value through the GP formula.
is there a easier way to solve it? esp since its the 3rd question in the test.
any easier way to solve this problem?
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t1 = 1/2
t2 = -1/(2^2)
Sn = a(1 - r^n)/(1 - r) = 1023/(1024*3) ~ 1/3
So it comes in between 1/3 and 1/2. Is it a way u were looking for?
t2 = -1/(2^2)
Sn = a(1 - r^n)/(1 - r) = 1023/(1024*3) ~ 1/3
So it comes in between 1/3 and 1/2. Is it a way u were looking for?
Correct me If I am wrong
Regards,
Amitava
Regards,
Amitava
I dunno whether this helps.
Term1: 1/2
Term2: -1/4
the sum is gonna be 1/2 - 1/4 +(x-x/2+x/4...)
the term in brackets cannot be more than 1/8 znd positive - because we are subtracting less and less with every new term.
so it is 1/4+y where (0<y<1/4)
so the sum is between 1/2 and 1/4
Term1: 1/2
Term2: -1/4
the sum is gonna be 1/2 - 1/4 +(x-x/2+x/4...)
the term in brackets cannot be more than 1/8 znd positive - because we are subtracting less and less with every new term.
so it is 1/4+y where (0<y<1/4)
so the sum is between 1/2 and 1/4
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Max,
This is the summation formula of GP (Geometric Progression).
Sn = a(1 -r^n)/(1-r) Where a = first term, n = no of terms and r = common ratio. Am I clear to explain?
This is the summation formula of GP (Geometric Progression).
Sn = a(1 -r^n)/(1-r) Where a = first term, n = no of terms and r = common ratio. Am I clear to explain?
Correct me If I am wrong
Regards,
Amitava
Regards,
Amitava
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GP means a series like - 2 + 2^2 + 2^3 + 2^4 + ...
Where according to my formula a = 2, r = 2^2/2 = 2 (Got it?) = Any term except the first one / previous term.
Am I able to clear ur doubt?
Where according to my formula a = 2, r = 2^2/2 = 2 (Got it?) = Any term except the first one / previous term.
Am I able to clear ur doubt?
Correct me If I am wrong
Regards,
Amitava
Regards,
Amitava
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Here's an easier way to look at this problem.
we have our formula ((-1)^(k+1))(1/(2^k))
the first part of the formula [(-1)^(k+1)]simply tells us if the resulting number will be positive or negative. For k=1 we get a positive 1. For k=2 we get -1. For k=3, +1, etc... If you notice for every odd value of k we plug in to the first part of the equation, we get +1. Likewise, for every even number we get -1. So this part of the equation is only there to change the sign of the entire equation.
I'm just going to call the other half [(1/(2^k))] of the equation above "1/eqk" for simplicity in the next part of my explanation.
The result of the first half will give us:
(1/eqk)-(1/eqk)+(1/eqk)-(1/eqk)+(1/eqk)-(1/eqk)+(1/eqk)+etc..
Now let's look at the second half of the equation : 1/(2^k)
let us plug in k=1. We get 1/2
For k=2 we get 1/4
k=3 we get 1/8
k=4 we get 1/16
etc...
if you notice it's simply asking for the inverse (1/x) of the first 10 powers of 2.
Most importantly, however, as we increase k the resulting number gets smaller.
Now let's look at the whole equation.
we now have
(1/2)-(1/4)+(1/8 )-(1/16)+etc... If you have caught on you will notice that we do not have to continue with the entire series to get answer.
1/2-1/4=1/4 now we add 1/8 (a smaller fraction than 1/4) and then subtract 1/16 (even smaller yet) from that. As k increase f(k) decreases.
We can now see that the resulting answer is bound somewhere between 1/4 and 1/2.
Thus the answer is D
we have our formula ((-1)^(k+1))(1/(2^k))
the first part of the formula [(-1)^(k+1)]simply tells us if the resulting number will be positive or negative. For k=1 we get a positive 1. For k=2 we get -1. For k=3, +1, etc... If you notice for every odd value of k we plug in to the first part of the equation, we get +1. Likewise, for every even number we get -1. So this part of the equation is only there to change the sign of the entire equation.
I'm just going to call the other half [(1/(2^k))] of the equation above "1/eqk" for simplicity in the next part of my explanation.
The result of the first half will give us:
(1/eqk)-(1/eqk)+(1/eqk)-(1/eqk)+(1/eqk)-(1/eqk)+(1/eqk)+etc..
Now let's look at the second half of the equation : 1/(2^k)
let us plug in k=1. We get 1/2
For k=2 we get 1/4
k=3 we get 1/8
k=4 we get 1/16
etc...
if you notice it's simply asking for the inverse (1/x) of the first 10 powers of 2.
Most importantly, however, as we increase k the resulting number gets smaller.
Now let's look at the whole equation.
we now have
(1/2)-(1/4)+(1/8 )-(1/16)+etc... If you have caught on you will notice that we do not have to continue with the entire series to get answer.
1/2-1/4=1/4 now we add 1/8 (a smaller fraction than 1/4) and then subtract 1/16 (even smaller yet) from that. As k increase f(k) decreases.
We can now see that the resulting answer is bound somewhere between 1/4 and 1/2.
Thus the answer is D