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OpenStudy (anonymous):
what's up?
OpenStudy (aravindg):
find sum upto n terms 8,88,888,8888.................
OpenStudy (nottim):
Help help help.
OpenStudy (nottim):
I like to help.
OpenStudy (nottim):
Me too.
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OpenStudy (anonymous):
Is it very hard ?
OpenStudy (aravindg):
more qns in que
OpenStudy (aravindg):
lol
OpenStudy (aravindg):
after this next qn
OpenStudy (anonymous):
what have you tried in this problem ?
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OpenStudy (aravindg):
taking 8 common
OpenStudy (anonymous):
and ?
OpenStudy (anonymous):
its all about GP
a= 8
r= 88/8=11
sum(n)= 8(11^n -1)
----------
10
OpenStudy (aravindg):
hey dashini i dont think r is 11
OpenStudy (anonymous):
This not a geometric progression
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OpenStudy (aravindg):
ya
OpenStudy (anonymous):
this question does involve a geometric series though (its hiding)
OpenStudy (aravindg):
how?
OpenStudy (anonymous):
The nth term is given by \( T_n=\large \frac{8\times(10^n-1)}{9} \)
OpenStudy (anonymous):
its a linear recursive function a(r+1) = 10*a(r) + 8
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OpenStudy (aravindg):
i need sm upto n
OpenStudy (anonymous):
Dhashni??? come to my post!
OpenStudy (aravindg):
sum
OpenStudy (anonymous):
something for which you should try using your head first ? :P
OpenStudy (anonymous):
\[8+88+888+8888+\cdots+888\ldots 88\]\[=8(1+11+111+1111+\ldots+111\ldots 11)\]
1111...111 is the same as:\[1+10+100+\ldots+ 10^{n-1}=\frac{10^n-1}{9}\]
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OpenStudy (aravindg):
hey joe u shud write n instead of 1
OpenStudy (aravindg):
isnt it???
OpenStudy (anonymous):
nopes joe wrote right.
OpenStudy (anonymous):
so like others have said, you are looking for:\[\sum_{n=0}^n8\cdot \frac{10^n-1}{9}\]
OpenStudy (anonymous):
8/9(10n\[8/9\ 10n+100-10^{n}\10]\]
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