Question

Find the sum to n terms of the series :
\[\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+.........\]

Answer 1

oh k so first of all it is an AP or GP or HP ... ? @shubham.bagrecha can u tell me?

Answer 2

i can be converted into A.G.P by taking 1/2 as x.

Answer 3

good .. so by taking 1/2 as x we have:
\[\large{x+ \frac{3x}{2} + \frac{7x}{4}+...}\]

Answer 4

right?

Answer 5

no

Answer 6

\[x + 3x^{2}+7x^{3}+15x^{4}+...............\]

Answer 7

hmn we can write like that also

wait ..

Answer 8

hint :\[a_n=\frac{2^n-1}{2^n}\]

Answer 9

this is for GP only ??

Answer 10

@mukushla ??

Answer 11

no this is the whole thing..

Answer 12

u can break it into\[\frac{2^n-1}{2^n}=1-(\frac{1}{2})^n\]

Answer 13

second part is GP

Answer 14

u can sum up it into n tems now

Answer 15

*terms

Answer 16

how this formula came?

Answer 17

can you explain step-wise.

Answer 18

\[\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+...=\frac{2-1}{2}+\frac{4-1}{4}+\frac{8-1}{8}+\frac{16-1}{16}+...\\=\frac{2^1-1}{2^1}+\frac{2^2-1}{2^2}+\frac{2^3-1}{2^3}+\frac{2^4-1}{2^4}+...\]

Answer 19

we are told to substitute 1/2 for x , then how will we solve?

Answer 20

for GP part right?

Answer 21

x+3x^2+7x^3+15x^4+...............

Answer 22

is this a GP ? this is not

Answer 23

how can we sum it into n term?

Answer 24

there ia a GP along with AP whose difference is also in GP

Answer 25

@hartnn

Answer 26

@TuringTest

Answer 27

i m on 2 other questions right now,after that i will look through this...

Answer 28

let me give u the formula for arithmetico-geometric sequence defined as:\[x_n=a_ng_n\] , where an and gn are the 'n'th terms of arithmetic and geometric sequences, respectively.

\[\frac{a_{n}g_{n+1}}{r-1}-\frac{x_{1}}{r-1}-\frac{d(g_{n+1}-g_{2})}{(r-1)^{2}} \]
or
\[ \frac{a_{n}g_{n+1}-x_{1}-drS_{g}}{r-1} \]
d is common difference of an and r is common ratio of gn,x1 if 1st term of your AGP
or equivalently u can use
\[t_{n}=\left[a+\left(n-1\right)d\right]r^{n-1}\]
and sum Sn will be:
\[S_{n}=\frac{a}{1-r}+\frac{dr\left(1-r^{n-1}\right)}{(1-r)^{2}}-\frac{\left[a+\left(n-1\right)d\right]r^{n}}{1-r}\]