Question

By Pmi prove that 1/2+1/4......1/[2^(n)]=1-1/[2^(n)]

Answer 1

induction ???

Answer 2

I have solved till p(k+1)= 1-1/(2^(k)) +1/(2^[(k+1)]

Answer 3

Ya induction.

Answer 4

solved for n=1
and n=k ????

Answer 5

Yes

Answer 6

for n=k what you got ??

Answer 7

1-1/2^(k)

Answer 8

and for k+1 what will we get ??

Answer 9

We should prove n=k+1 is 1-1/[2^(k+1)]

Answer 10

first put n=k+1

Answer 11

we have to bring left side equal to right side

Answer 12

put n=k+1 on left

Answer 13

it will form a geomatric progression

Answer 14

Ya.... The lhs is 1-1/(2^(k)) +1/(2^[(k+1)].... I have to prove that this is equal to1-1/[2^(k+1)].

Answer 15

1/2 +1/4 + 1/8 +1/16+1/32+..................

Answer 16

upto k+1 term

Answer 17

Ya... That would be 1/2^(k+1)

Answer 18

aum has also a good approch (if dont know abt GP)

Answer 19

I know GP. And aum's approach is what I generally do. Was confused with solving the exponents.

Answer 20

Assume it is true for k. \(\Large p(k) = 1 - \frac{1}{2^k} \)
The \((k+1)^{th}\) term is: \(\Large \frac{1}{2^{k+1}}\)
Therefore, \(\Large p(k+1) = p(k) + \frac{1}{2^{k+1}}\)
\(\Large p(k+1) = 1 - \frac{1}{2^k} + \frac{1}{2^{k+1}} =
1 - \frac{1}{2^k} + \frac{1}{2~*~2^{k}} = \\ \Large
1 - \frac{1}{2^k}(1 - \frac 12) = 1 - \frac{1}{2^k}\frac 12 = 1 - \frac{1}{2^{k+1}}
\)
If it is true for k, it is true for (k+1).
And it is true for n = 1 and therefore it is true for all n.

Answer 21

Thanks a lot @aum

Answer 22

You are welcome.