Question

Can anyone explain Method of difference of GP

Answer 1

@ganeshie8 @hartnn pls help

Answer 2

@ikram002p hlp

Answer 3

never heard of that....what is it ?

Answer 4

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Answer 5

In my text book it is written as a trick to solve question
In difference GP
Tn=ar^n+b where r is the common ration

Answer 6

like if u have
\(S1= a+ar+ar^2+.....ar^n\)

so find difference btw two terms

Answer 7

it means when a series is given it's difference is in GP

Answer 8

well mmm dint uderstand the Qn XD
and sadly im going for dinner :'(

Answer 9

That is so cool ikram how did you do tht?!

Answer 10

@camerondoherty ikr :3

Answer 11

LAtex has so many things behid it and secret it so cool

Answer 12

http://www.askiitians.com/iit-jee-progressions-and-series/method-of-differences/

Answer 13

@hartnn i have a question where series 1,3,7,15....
so the difference is in gp 2,4,8................

Answer 14

@ParthKohli hlp

Answer 15

ok, so you want to find the sum of series 1,3,7,15...
using this method ?

Answer 16

the nth term

Answer 17

lets see, even i am doing it for first time

S = 1+3+7+15+....tn
S = 1+3+7+15+....+t(n-1)+ tn

Subtracting these

0 = 1+ [2+ 4+8+...] + t_n

Answer 18

yes

Answer 19

0 = 1+ [2+ 4+8+...] -t_n
actually

Answer 20

yes tn-1=2,4,8

Answer 21

............

Answer 22

so t_n = 1+ S_(GP)

S_(GP) is sum of GP series 2,4,8,...

Answer 23

yes

Answer 24

same steps but i want it in terms of n

Answer 25

\( t_n = 1+ 2 \dfrac{2^n-1}{2-1} = 1+ 2^{n+1}-2=2^{n+1}-1 \)
^^^ in terms of n

Answer 26

ohh yes

Answer 27

here n starts from 0 ! :O

Answer 28

\[2^{n+1}-1/2^n\]

Answer 29

why?

Answer 30

just pls tell

Answer 31

we got t_n = 2^(n+1) -1, n=0,1,2,3...

for 1,3,7,15...actual t_n is 2^n-1, n =1,2,3...

Answer 32

yes so my answer was right thnx for ur hlp @hartnn

Answer 33

ok, welcome ^_^