IIT JEE , AIEEE Forum - Mathematics , Algebra

DON007

everybofy is discussing 3rd one & 2nd one.........

i should discuss the 1st one....isnt it...........now....

look...........

abc= 30.......

triplets can be formed is as follows

1) 6 5 1 = 3!

2)15 1 2 = 3!

3) 2 3 5 = 3!

4) 3 10 1 = 3!

add them we get................

= 24 .........(ans).

hope i m correct..............

thanku............

and all r right .....pakka..........

in third one p+q=2....

now 4 second prblm............

(1+x+x²)²⁰=a₀+a₁x+...........

differnetiate both sides w.r.t. x to get

20(1+x+x²)¹⁹(0+1+2x)=0+a₁+2a₂x+3a₃x²+.....

multiply by x we get

x *20(1+x+x²)¹⁹(0+1+2x)=0+a_1^x+2a₂x²+3a₃x³+.......

now put...........

x= - 1

we get........

-a₁+2a₂-- 3a₃+.......+40a_{40 ^{= 20.......}}

freniend and ankur has done a good job .............

but first multiply by x again............

okkkkkkkk

........ thanku..........

hope u understand.........