Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion

Algebra

Forum Expert

Joined:	18 Dec 2007
Post:	926

11 May 2008 12:54:58 IST

0 People liked this

801

Challenge in Algebra - From Feynmann

None

$For\;any\;positive\;integer\;n\;let\;<n>denote\;the\;$

$integer \;nearest\;to\;\sqrt{n}$

$Show\;that\;$

$\sum_{n=1}^{inf}\;({2}^{<n>}\;+{2}^{-<n>})/{2}^{n} =3$

All experts are welcome !!!

Share this article on:

Comments (7)

Rahul

Blazing goIITian

Joined: 13 Jul 2007

Posts: 547

11 May 2008 16:34:57 IST

Like 3 people liked this

Let t_r be the rth term of the series.

t1 = (2¹+2^-1)/2¹; t2 = (2¹+2^-1)/2²................2 terms with (2¹+2^-1)

t3=(2²+2^-2)/2³; t4=(2²+2^-2)/2⁴likewise t5 and t6............4 terms with (2²+2^-2)

t7=(2³+2^-3)/2⁷; t8 =(2³+2^-3)/2⁸so on upto t12......... 6 terms with (2³+2^-3)

Likewise there will be 2r terms with (2^r+2^-r)

Let, T₁=t1+t2 = [(2¹+2^-1)/2¹]* (1 + 1/2)

T₂= t3+t4+t5+t6 = [(2²+2^-2)/2³]* (1+1/2+1/2² + 1/2³)

T₃= t7 + t8 + ... + t12 = [(2³+2^-3)/2⁷]* (1+1/2 + 1/2² +....6 terms)

T_r= [(2^r + 2^-r) / 2^{r^2-r+1}] * (1 + 1/2 + 1/2² + .... 2r terms)

= [(2^r + 2^-r) / 2^{r^2-r+1}] * 2(1-2^-2r)

Simplifying the above expression, we get

T_r = 2^{2r-r^2} - 2^{-2r-r^2}

Now,

T1 = 2¹ - 2^-3

T2 = 2⁰ - 2^-8

T3 = 2^-3 - 2^-15

T4 = 2^-8 - 2^-24

T5 = 2^-15 - 2^-35

--- --- ----

upto infinity...

Adding up the above terms, we get S = 2¹ + 2⁰ (other terms getting cancelled out)

S=3

(sory cudn't shorten it much :( )

Rahul

Blazing goIITian

Joined: 13 Jul 2007

Posts: 547

11 May 2008 16:35:31 IST

Like 0 people liked this

pls point out in case of any error

Gaurav |spideyunlimited| Ragtah

Moderator

Joined: 16 Dec 2006

Posts: 3373

11 May 2008 17:20:34 IST

Like 0 people liked this

i think ur right
i was going along the same lines of
2 terms having same, then next 4 terms, then 6 terms :)

Mukund

Blazing goIITian

Joined: 17 Apr 2008

Posts: 432

11 May 2008 17:58:28 IST

Like 0 people liked this

Yeah.. I think you're right..

abhishek sinha

Forum Expert

Joined: 18 Dec 2007

Posts: 926

12 May 2008 19:20:24 IST

Like 0 people liked this

Hey , I must say that it is a good observation , but u have to provide the complete proof ..........

'Likewise there will be 2r terms with (2^r+2^-r)' prove it mathematically !!

Rahul

Blazing goIITian

Joined: 13 Jul 2007

Posts: 547

12 May 2008 20:43:52 IST

Like 4 people liked this

Wel, the number of terms with (2^r + 2^-r ) = the number of integers 'n' where with <n> = r.

Let, r + 0.5 > $\sqrt{n}$ >= r - 0.5

where r is an integer. For all such n, <n> = r

r² + 0.25 + r > n >= r² + 0.25 - r.

For n to be an integer,

r² + r >= n > r² - r

Obviously the number of such integers 'n' = 2r

abhishek sinha

Forum Expert

Joined: 18 Dec 2007

Posts: 926

12 May 2008 22:36:17 IST

Like 0 people liked this

good !

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team

Free Sign Up!

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor

@ INR 1,995/-

For Quick Info

Find Posts by Topics

Physics.

Topics

9389

20086

3671

2185

7614

2826

2634

Mathematics.

Chemistry.

Biology

Institutes

Parents

Board

Fun Zone