Question

1/2 + 1/6 + 1/12 + 1/20 +… + 1/9900 =

Answer 1

@mathmate @brainzonly

Answer 2

\(\Large \sum_{i=1}^{99} \frac{1}{i(i+1)}\), right?

Answer 3

Just need to find a sequence that describes
2, 6, 12, 20, 30, 42
+4, +6, +8, +10, +12
+2, +2, +2

By eyeballing it, it's \[\large n^2 + n\]
and for 9900 = n^2+n, the last value is n=99

So your series is \[\large \sum_{n=1}^{n=99} \frac{ 1 }{ n^2 + n }\]

Answer 4

If so, then use
\(\Large \frac{1}{i(i+1)}\equiv\frac{1}{i}-\frac{1}{i+1}\)
and sum the truncated harmonic series.

Answer 5

that's very complicated :/
Can you explain me further or use easier way?

Answer 6

hint: read about telescopic series.

Answer 7

ok

Answer 8

Which part don't you understand?

Answer 9

You look at the denominator, and you will find that 2,6,12 ... ... Can be obtained by multiplying two consecutive natural numbers.



You can type turns into the original



1/1*2+1/2*3+1/3*4+1/4*5+......+1/99*100=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+......+1/99-1/100

Parts are found in the middle of all the simplicity of the AH? Above I have no parentheses, bored, you can read it.

And =1-1/100=99/100

The answer is 99/100.

Later this topic to learn how to analyze and question you must be characteristic of the flexibility of thinking to analyze and understand, of course, long after will naturally think of the method.

Answer 10

@agent0smith how do you know 9900 = n^2+n, the last value is n=99

Answer 11

By solving for n.\[\large 9900 = n^2+n\]\[\large 0 = n^2+n - 9900 = (n+100)(n-99)\]

Answer 12

Owhh.. I see

Answer 13

Then?

Answer 14

See @brainzonly's method, it's an easy way to find the sum.

Answer 15

@brainzonly

How do you get it's same with 1 - 1/100 ?

Answer 16

The sum breaks down to
\(\Large \sum_{n=1}^{100} \frac{1}{n}-\frac{1}{n+1}\)
=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)....(1/98-1/99)+(1/99-1/100)
[note that the middle terms cancel out, in a telescopic series]
=1-1/100
=99/100

Answer 17

I got it.
Thx guys!

Answer 18

Using what @mathmale and plugging in the first few values of i=1, 2, 3 should show it to you...
\[\large \frac{1}{i}-\frac{1}{i+1} =\left( \frac{1}{1}-\frac{1}{1+1} \right) + \left( \frac{1}{2}-\frac{1}{2+1} \right) +\left( \frac{1}{3}-\frac{1}{3+1} \right) +...\]
\[\large \frac{1}{i}-\frac{1}{i+1} =\left( \frac{1}{1}-\frac{1}{2} \right) + \left( \frac{1}{2}-\frac{1}{3} \right) +\left( \frac{1}{3}-\frac{1}{4} \right) +...\]

Answer 19

@mathmate not mathmale

Answer 20

@agent0smith :) we're so confusing!

Answer 21

That's why everybody likes ME. :D @mathmate is probably smarter though.