4
UTTARA
·2009-11-08 23:20:41
S = 1 / 30+0 + 1/3(1)+(1) +.............+∞
= 1/(1-1/32 )
= 9/8
62
Lokesh Verma
·2009-11-08 23:22:48
Uttara you have solved the other one.. where i=j
;)
4
UTTARA
·2009-11-08 23:32:20
Sorry I mistook the qs
Any hint for the right approach ??
62
Lokesh Verma
·2009-11-08 23:35:00
yeah..
What you have done is a subset of the work needed :D
3
msp
·2009-11-09 00:30:13
Reqd sum=9/64 is this rite.
341
Hari Shankar
·2009-11-09 00:59:43
to get some idea on how to proceed, look at the derivation of the sum of divisors of a number.
62
Lokesh Verma
·2009-11-09 01:13:27
no sankara.. you have missed it by a whisker
1
Anirudh Kumar
·2009-11-09 02:27:33
sir is
S= 9/4 correct .
using sum of infinite terms of a G.P
keeping i =0 varying j from o to ∞
keepint i=1 and varying j from o to ∞
i.e
S= 1/30{ 1/30+1/31+...+1/3∞}
+1/31{ 1/30+1/31+...+1/3∞}
+...
62
Lokesh Verma
·2009-11-09 02:35:56
anirudh you have also missed out something.. see more closely..
1
Anirudh Kumar
·2009-11-09 02:43:52
sir
i≠j .
thus from my solution i have to subtract those cases of
i=j
62
Lokesh Verma
·2009-11-09 02:55:16
yup.. so the final answer is?
1
riya
·2009-11-09 03:00:31
the answer is 3/4 am i correct
106
Asish Mahapatra
·2009-11-09 03:01:52
if anirudh's calc is correct,
the final ans is S - [(1/30)2 + (1/31)2 + ...]
= 9/4 - 11-1/9
= 9/4 - 9/8
= 9/8
verification of anirudhs calc
S = (1/30 + 1/31 + 1/32 + ... )2
= (11-1/3)2
= 9/4
1
Anirudh Kumar
·2009-11-09 03:02:42
S =
9/4-9/8
= 9/8
sir this answer is same as @uttara answers so i got a little Confused
62
Lokesh Verma
·2009-11-09 03:03:05
There are some more proofs.. can you think of a couple more?