211st term will be zero.....
2nd term socjna padega
(1+x)10 = a0 + a1 x +a2 x2 + ..... + a10 x10
then,
(a0 - a2 + a4 - a6 + a8 - a10)2 + (a1 - a3 + a5 - a7 + a9)2
is equal to : ??
-
UP 0 DOWN 0 0 11
11 Answers
1oh bhaiyaaaaa u are great!!!
yes got it!!
took one x=i and another x=-i and multiplied ....
wow!!!!!!!!!
1tell me on thing... how did this 'i' thing came to your mind ???
wow !
its simply sweet !!!
1357putting x=i you will get
(1+i)10=(a0-a2+a4-a6+a8-a10)+i(a1-a3+a5-a7+a9)
takiking modulus
(a0-a2+a4-a6+a8-a10)2+(a1-a3+a5-a7+a9)2=(|1+i|2)10=210
why take separate cases if it can be solved in one
1357hehe
first I was trying to put 1 or -1, then iota came in my mind
1no actually i din see your post .. i thnk u edited later...
firstly wen u wrote take x=i .. i took it .. and then
since i am a bit stupid ... that mod thing din came to mind
and i took x=-i and multiplied ....
but great bhaiya!
11As a backdrop to this one i recommend everyone to look into this one
http://targetiit.com/iit_jee_forum/posts/great_one_2452.html
similar logic