All experts are welcome !!!

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)/2⁴ likewise t5 and t6............4 terms with (2²+2^-2)

t7=(2³+2^-3)/2⁷; t8 =(2³+2^-3)/2⁸ so 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 :( )

Yeah.. I think you're right..

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 !!

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

Let,    r + 0.5 > >= 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

good !