Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- shubhamsrg

infinite summation :
1 + (1/3) + (1.3 / 3.6) + (1.3.5 / 3.6.9) + (1.3.5.7 / 3.6.9.12) ........ = ?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- shubhamsrg

infinite summation :
1 + (1/3) + (1.3 / 3.6) + (1.3.5 / 3.6.9) + (1.3.5.7 / 3.6.9.12) ........ = ?

- schrodinger

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- shubhamsrg

so far, my progress has been this much :
i started my series from (1/3) ...
denom. of nth term is easily (3^n)*(n!)
num. of nth term will be
1.3.5.7............
=(1.2.3.4.5...........)/(2.4.6.8........)
=( (2n)! )/(2^n * n! )
thus nth term will be
(2n)! / (n!)^2 * 6^n
or C(2n,n) / 6^n
we have to fine its summation from n = 1 to infinity..
but am clueless after this..
will this be even helpful ?

- experimentX

that's central binomial coefficient ... google for something like it.

- shubhamsrg

well i have googled over,,i got this solution,,but couldnt understand as to why we did that..
it said to compare the summation with (1+x)^n
and solve ultimately,,and to my surprise, i got the ans.. if anyone can explain why ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- experimentX

it's better to use the generating function directly
http://mathworld.wolfram.com/images/equations/CentralBinomialCoefficient/NumberedEquation1.gif

- shubhamsrg

didnt get you ?

- experimentX

http://mathworld.wolfram.com/CentralBinomialCoefficient.html
put x=1/6 and get it's value

- shubhamsrg

ohh,,nice !! where does this come from ?

- experimentX

|dw:1349165062856:dw|

- shubhamsrg

but its C(2n,n) ?!?

- experimentX

yeah ... I'm just playing around. I've seen that method too ... kinda forget it though.

- shubhamsrg

hmm,,so i should just rote it for the time being !! hmm..anyways ,,thank you.. :)

- shubhamsrg

should i just tag @mukushla @eliassaab

- experimentX

|dw:1349165603145:dw|

- experimentX

probably it comes from what google said
\[ (1+ax)^k = \sum_{i=0}^k\binom{k}{i}(ax)^i = \sum_{n=0}^\infty \binom{2n}{n}x^k\]
since it's infinite k can't be natural number.

- experimentX

have to prove that a=-4 and k=-1/2

- experimentX

here is one
\[ 1 + {1 \over 3} + {1 \cdot 3 \over 3\cdot 6} + {1\cdot3\cdot 5\over 3\cdot6\cdot9}+...\]
\[ (1 + x)^n = 1 + nx + {n(n-1) \over 1 \cdot2} x^2 + ....+\infty \]
assuming n to be non natural
\[ nx = {1 \over 3}\\
{n(n-1)\over 1 \cdot2}x^2 = {1 \over 6} \\
\text{ this gives }
x={-2 \over 3} , n={-1 \over 2}\]

- experimentX

\[ \left(1 - {2 \over 3}\right)^{-1 \over 2} = \sqrt{3}\]

- shubhamsrg

hmm..i get the idea i guess.//thanks !! :)

- experimentX

there is bigger idea
http://math.stackexchange.com/questions/205898/how-to-show-that-1-over-sqrt1-4x-generate-sum-n-0-infty-binom2n

- shubhamsrg

ohh yes,,gotcha,,that was really helpful..thanks a ton sir!

Looking for something else?

Not the answer you are looking for? Search for more explanations.