how 2 solve dis...??

1+(1+(1/2))+(1+(1/2)+(1/4))+.......to n terms

plz ...plz ...tell me dis..

Dear srishti

sum of nterms can be written as  as n+n-1/2 + n-2/4-----+1/2n-1

which is sum of nterms of an AGP

The sum Sn of first n terms of an A.G. P. is obtained in the   following way :

 

Sn = ab + (a + d)br + (a + 2d)br2 +.........+(a + (n - 2)d)brn-2 + (a + (n - 1)d)brn-1
 

Multiply both sides by r, so that
 

r Sn = abr + (a + d)br2+.........+(a + (n - 3)d)brn-2 + (a + (n - 2)d)brn-1 + (a + (n - 1)d)brn
 

Subtracting, we get
 

(1 - r)Sn = ab + dbr + dbr2 +.......+dbrn - 2 + dbrn - 1 - (a + (n - 1)d)brn
 

= ab + dbr/(1–rn–1)/(1–r) – (a+(n–1)d)brn => Sn = ab/1–r + dbr(1–r^n–1)/(1–r)2 – (a+(n–1)d)brⁿ/1–r
 
a=n

b=1

r=.5

d=-1

So Sn = 2n+(-1)*0.5*1(1-0.5^n-1)*4-2ⁿ

=>Sn = 2n-2(1-0.5^n-1)-2ⁿ

Please feel free to post as many doubts on our discussion forum as you can.we will get you the answer and detailed  solution very  quickly.

All the best.

Regards,
Askiitians Experts
Prudhvi Teja

Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian