Question

Find the sum to n terms \[\frac{ 1 }{ 1*2 }+\frac{ 1 }{ 2*3 }+\frac{ 1 }{ 3*4 }\]

Answer 1

find the general terms ... make partial fractions .. then use telescoping.

Answer 2

WAT IS TELESCOPING

Answer 3

first do the first two steps

Answer 4

i got \[\frac{ 1 }{ n(n+1) }\]

Answer 5

then do the second step.

Answer 6

Hw to do that

Answer 7

don't you know about partial fractions?

Answer 8

\[\frac{ A }{ n }+\frac{ b }{ n+1 }\]

like this

Answer 9

yeah .. A=1 and b=-1

Answer 10

\[\frac{ 1 }{ n }-\frac{ 1 }{ n+1 }\]

Answer 11

better google for video on Heaviside cover up method for partial fractions. it's not easier.

Answer 12

not -> *lot

Answer 13

i knw....partial fraction

Answer 14

then nw

Answer 15

add the series.

Answer 16

Sn = n(n+1)/2

Answer 17

that's harmonic series ... you can't do that.

Answer 18

Oh.....

Sn = a1+a2........an

Answer 19

put n=1, 2, 3, ... and evaluate the sum.

Answer 20

a1 = 1/1 - 1/2

a2=1/2-1/3.......................... an = 1/n -1/n+1

Answer 21

yeah yea ... put those up and cancel the + - and get your final result
1 - 1/n+1

Answer 22

1/1 - 1/n+1 will remain

Answer 23

Yup..) thxxx..)

Answer 24

\[\frac{ 1 }{ 1\times2 }+\frac{ 1 }{ 2\times3 }+\frac{ 1 }{ 3\times4 }+......\]\[=\frac{ 2-1 }{ 1\times2 }+\frac{ 3-2}{ 2\times3 }+\frac{ 4-3 }{ 3\times4 }+......\]\[=1-\frac{ 1 }{ 2 }+\frac{ 1 }{ 2 }-\frac{ 1 }{ 3 }+\frac{ 1 }{ 3}-\frac{ 1 }{ 4 }+\frac{ 1 }{ 4}-\]\[=1\]

Answer 25

It is to infty.