Math Is Fun Forum

It should be7n*{1/(280n)-1/(8*140^2*n^2)}=1/(40)-7/(8*140^2*n)

v2=0 m/s (Stopping at rest)
u2= v m/s
a2=- 3a m/s²

v²=u²+2as
⇒s=(v²-u²)/2a
⇒s2=(v2²-u2²)/2a2
⇒s2=(-u2²)/2a2
⇒s2=(v²)/6a
s1+s2=v²/2a+v²/6a=2v²/3a

I like clarity on this point.Is it not horizontal distance between opposing windows?And what is the value of 'g' in fps units? I vaguely remember it as 32 ft/sec^2. Is it right? I am used to S.I. units.

It comes readily from the series expansion of log(1+x) and log(1-x)

For the new problem get u in terms of gt and calculate s1 and s2. Simple. I am afraid I should not reveal anything more.

That trial and error is quite easy because of 5.e.g. 173 You have to find a multiple of 17 ending with 3 or 8 i.e 68+105 or 153+20  but the resulting i and j you have to check for the condition.

hi Monox D. I-Fly,
It was just for fun.I have neither  objection ,nor any authority nor any intention to do so. My contention was it may not benefit the one who asked the question. I too have responded to queries 2 to 3 years back when the topic was found interesting.

When I got this  182*{-97*929}-165*{-107*929}=929 I was not happy since i and j were negative but since your condition was on integer I allowed it.However it is possible to add and subtract packets of 182*165 as many as you like to make them positive keeping eye on relatively prime condition. I added and subtracted 552(547 is sufficient to make them positive but relatively prime condition could not be met) packets to each term.Thus 182*{-97*929+552*165}=967 and 165*{-107*929+552*182}=1061. as I told you I relied on only (1) and (2) in #3 but more such relations could be found.

mr.wong,
I am very sorry,I had not read the problem carefully. 182 and 165 are relatively prime but I had overlooked that for i and j. I was wondering why there are only some specific numbers on R.H.S.As a matter of fact I had not gone through your earlier thread on prime numbers.