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Uploaded by mlehain on Mar 13, 2009
This is all about how we can come up a with a formula to calculate the SUM of a geometric series. Clever stuff...
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WHERE IS THE NEXT VIDDDDDD
MikSane 7 months ago
were is the next video
youdanaregood 9 months ago
but why would you multiply it by r in the second place?
joielock 9 months ago
AWSOME!!!...... may i suggest u to get a White board..... and some erasable markers..... so u dont have to buy Pages all the time...... just write and erase...... the viewers can re watch the video if they miss something.........
Taimoorabdullah 9 months ago
is that "r" the pro-numeral that is written below the sigma??
poyanator 10 months ago
saviour!
vesaab 1 year ago
Welll.. with this info i was able to find the sum of un = (-1)^n . (3^n/pi^(n+1))...
Thx a lot sir.. ;) .
One bit closer to getting an A on that final exame of Analysis.
BrightnessKnight 1 year ago
Thanks. You explain things so clearly.
LikeItDeep 1 year ago
when you are doing the sum, you did
s=a+ar+ar^2+ar^3+.........+ar^n-1 and multiplied by r, which gave u
rs= ar+ar^2+ar^3+..........+ar^n but had you continued s to ar^n and multiplied it by r, you would have gotten ar^n+1
utubsdum 1 year ago
Brilliant, I wish you would cover more maths topics. It's so much easier to follow on here than in the lecture. :D
MrDrunklove 1 year ago
Load more suggestions
WHERE IS THE NEXT VIDDDDDD
MikSane 7 months ago
were is the next video
youdanaregood 9 months ago
but why would you multiply it by r in the second place?
joielock 9 months ago
AWSOME!!!...... may i suggest u to get a White board..... and some erasable markers..... so u dont have to buy Pages all the time...... just write and erase...... the viewers can re watch the video if they miss something.........
Taimoorabdullah 9 months ago
is that "r" the pro-numeral that is written below the sigma??
poyanator 10 months ago
saviour!
vesaab 1 year ago
Welll.. with this info i was able to find the sum of un = (-1)^n . (3^n/pi^(n+1))...
Thx a lot sir.. ;) .
One bit closer to getting an A on that final exame of Analysis.
BrightnessKnight 1 year ago
Thanks. You explain things so clearly.
LikeItDeep 1 year ago
when you are doing the sum, you did
s=a+ar+ar^2+ar^3+.........+ar^n-1 and multiplied by r, which gave u
rs= ar+ar^2+ar^3+..........+ar^n but had you continued s to ar^n and multiplied it by r, you would have gotten ar^n+1
utubsdum 1 year ago
Brilliant, I wish you would cover more maths topics. It's so much easier to follow on here than in the lecture. :D
MrDrunklove 1 year ago