Question

the sum of the series 20c0 - 20c1 + 20c2 - 20c3+........+20c10 is

Answer 1

20(c0-c1+c2-c3 + ... + c10); what are the coeefs?

Answer 2

i think it should use binomial form (1-x)^20 or somewhat like it...

Answer 3

no no no hold on.. its a combinatorics problem. the 'c' here means that

Answer 4

use (1-x)^20+(1+x)^20 you will get the ans...

Answer 5

deeprony have u understood the p problem??????

Answer 6

20!/20! - 20!/19! + 20!/18!2! .... like that?

Answer 7

"(1-x)^20+(1+x)^20" what is x here??

yea i understood that quetion=)

Answer 8

yes mistre thats correct

Answer 9

i stil gotta do these the brute force way ;)

Answer 10

isint there a formula??

Answer 11

(1-x)^20=20c0-20c1x+..........
now 20cn=20c20-n.
and thus 20c20=20c0.
and so on
put x=1.
20c0-20c1+.....=0
2(20c0-20c1-.......+20c10)-20c10=0
hence ans is (20c10)/2

Answer 12

can you explain what you did hre.. 20c0-20c1+.....=0
2(20c0-20c1-.......+20c10)-20c10=0

Answer 13

i put just x=1 in the 1st expansion of (1-x)^20

Answer 14

yea ok wat did u do thn??

Answer 15

that is the way of doing.....using a binomial series....

Answer 16

i had (1-x)^20=20c0-20c1x+20c2x^2....
put x=1
0=20c0-20c1+20c2......

Answer 17

use 20c20=20c0,20c19=20c1,.......20c11=20c9.

Answer 18

i got that olryt... i didnt understnd what u did xt

Answer 19

*nxt

Answer 20

you get the term 20c0,20c1,....20c9 twice but 20c10 only one...

Answer 21

ahh i c... got it now...=))

Answer 22

thus i write ..
0=2(20c0-20c1+20c2........+20c10)-20c10

Answer 23

okkkkkkkkkkkkkkkkkk?????????

Answer 24

yea!

Answer 25

these are some tricks to do it...............

Answer 26

is the answer 0