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i and j are vectors along x and y axis resp. What are the components of vector
A = 2i + 3j along the directions of i+j and i - j ..................

let me try this one

now got it ....
and when n>x>a

the general term is ⁿC_rx^n-ra^_rthe coefficient of all the terms after n=x will be more than xⁿ
Here we are concerned with the coefficient only:
ⁿC_r= n!/[(n-r)!r!]

therefore the largest term will be that term where r satisfies or nearly satisfies the following condition:
(n-r)!r!=x!

do rate me if i am correct

ya the above one works only if n<x>a

yes quite tricky it is !

ya the third term
ⁿC₂ 100 x 7²⁵⁴= 256 x 255 x 100 x 7²⁵⁴
=6528000 x 7²⁵⁴
so no question of considering it as there are 3 xeros

plz do rate me if i am right

here is the proof
we can expand (7+10)²⁵⁶ binomially

since a= 10 every other term other than the first and the second will have more than 2 zero.

so lets see the first term
ⁿC₀7^₂₅₆₌7²⁵⁶the pattern goes like 7 to the power 1k,5k,9k .......(4n-3) end with_07
                               7 to the power 2k,6k,10k ......(4n-2) end with _49
                               7 to the power 3k,7k,11k ......(4n-1) end with_43
                               7 to the power 4k,8k,12k ...... (4n-0) end with_01
Thus 7^256 must end with _01............. again by recognising the patttern i came to a conclusion the the last three digits are 601

now the second term:
ⁿC₁7²⁵⁵10 = 2550 x 7²⁵⁵
the last three digits of this will be 850

ADDind the two 850+601 we get the last digits to be 451

No it must end with 451 ......... if this is right i will post the proof.

well i dont know soo well nut here is my try at this

If (x+a)ⁿis the expression we can expand it thru binomial theorem there are three possibilities:

if x>a:
then
ⁿC₀xⁿwill be the highest

if x=a:
ⁿC_n/2x^n/2a^n/2 will be the highest

DIng dong ding dong DIng kukukuku its 9.30!!!!! Answer n proof plzz..............

This was a tough one!! waiting to see the answer ...........

is the general term

^2r+1 C _r 2^-3r

its good that you gave hints !!! but seriously its blowin my head off!!

wow atlast the answer came!

good work amit!

OK

_{^{Let the series be a,a+d,a+2d .................}}

when n=7

S_{7 ^{< 0}}

_{^{7/2 ( 2a+6d) < 0}}

_^a+3d<0

_{^{means the 4th term < 0}}

_{^{when n= 11}}

_^S11>0

_{11/2(2a+10d)>0}

_a+5d>0

_{the 6th term >0}

_{It means that till 4th term evry term must be -ve n after 6th term +ve . 5th cud be anything.}

_{thus a must be some -ve number.}

_{if 5th term is = 0}

_{then max no of terms = 10}

_{if 5th term>0}

_{max no of terms =9}

_{if 5th <0}

_{max no of term= 11}

_{since those 11 cud be anywhere in the series}

_{max no of terms = 12}

_{if it is more than that the fourth term will become more than 0}

_{thus answer =12}