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# how 2 solve dis...??1+(1+(1/2))+(1+(1/2)+(1/4))+.......to n termsplz ...plz ...tell me dis..

Prudhvi teja
83 Points
11 years ago

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–rn–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(1-0.5n-1)*4-2n

=>Sn = 2n-2(1-0.5n-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,
Prudhvi Teja

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