4Thanks Nishant S i r & Anant Sir : )
In (1 + x + x2 + 2x3 + x4 + x5 + x6)n the coefficient of x4 is??????
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23 Answers
1Dont think much directly use multinomial theorem to get the ans straight away
3wat are u saying ............ expand the answer ive given ull get the same thing wat anant sir got ........ i donno wat ur arguin about ... !!!!!
4@iitimcoming The qs is
In (1 + x + x2 + 2x3 + x4 + x5 + x6)n the coefficient of x4 is??????
Put n=1
We get coeff of x4 as 1
Coming to "We donno whether this condition is tru here
as we donno n""" wat are u talkin abt .. this is basic binomial stuff ......its always true ........
Comparing ur formula and the qs we get m = n , r1 =4
But we donno whether m=n=4
So I think that can't be used
3lets take a simple expansion (1+x)^n ....... every term will have (1)^r (x)^(n-r)
the sum of thier powers is always ... n!!!!!!!!!!!! thats all there is to it [r1 + r2 ...... = m]
3and abt that r1 + r2 ...... = m ........."""We donno whether this condition is tru here
as we donno n""" wat are u talkin abt .. this is basic binomial stuff ......its always true ........
3if n =1 .... then we cannot have any tem where the term x[(1 + x + x2 + 2x3 + x4 + x5 + x6)n] is raised to the power 4 ...... !!!!!!!!!! so that particular term will not exist ..... im assumin n>=4 .....
4No I think there's some mistake in ur solution
If n=1 How will we compute -3 !
See ur formula is correct but r1 + r2 +...= m
We donno whether this condition is tru here
as we donno n
3(1)^a (x)^b (x^2)^c (2x^3)^d (x^4)^e
so here to get x^4 we have
(i) b=4 , c=d=e=0 , a=n-4
(ii) e=1 , b=c=d=0 , a=n-1
(iii) b=1 , d=1, c=e=0 , a = n-2 .....
so we have ...
coeff of x^4 =
n!/(n-4)!4! + n!/(n-1)! + 2n!/(n-2)! .........
3a nice result ...... comes handy durin exams where time is important ,,,,
in the expansion
(x1+x2+x3+x4+x5.....xn)^m
where n,m==> N ......
the coeff of x1^r1 . x2^r2 . x3^r3 ............ xn^rn
where r1 + r2 +r3 ... rn = m
is
m!/(r1! r2! .......rn!) ...................
24(1 + x + x2 + 2x3 + x4 + x5 + x6)n
= (1 + x + x2 + x3 + x3 + x4 + x5 + x6)n
= (1 + x + x2 + x3)n (1 + x3)n
= (1 + x)n (1 + x2)n (1 +x3)n
= [1 + C1x + C2x2 + C3x3 + C4x4 +..][C0 + C1x2 + C2x4 + ..] [C0 + C1x3+... ]
I tried to do this way today..
is ths OK ???
21(1+x)n (1+x2(1+x))n
now general term of (1+x)n is nCr xr
general term of (1+x2(1+x)) is nCr x2r(1+x)r
to find coefficient of x4 it will be
(coefficiet of x0 in (1+x)n * coefficiet of x4 in (1+x2(1+x))n +
coefficiet of x in (1+x)n * coefficient of x3 in (1+x2(1+x))n +
coefficient of x2 in (1+x)n * coefficient of x2 in (1+x2(1+x))n + coefficient of x3 in (1+x)n * coefficient of x in (1+x(1+x))n + coefficient of x4 in (1+x)n * coefficient of x0 in (1+x2(1+x))n)
that comes out to be
nC0. nC2 +nC1 .nC1 +nC2. nC1 +nC4. nC0.........
66Well I forgot to the divide by 24. (I was feeling a bit sleepy; I posted the answer at around 4 in the morning[1]).
The correct answer should be
n(n3 + 6 n2 + 35 n - 18) 24.
One of the ways (which is very general) to obtain this is as follows:
Let
f(x)=(1+x+x^2+2x^3+x^4+x^5+x^6)^n
Suppose we expand f(x) in Taylor series around the origin. Then the coefficient of x4 will be
\dfrac{f^{(4)}(0)}{4!}
Therefore, this must be the coefficient of x4 in the expansion of f(x).
But for the present problem, this is unnecessary. As Nishant Sir already pointed out, the required coefficient is the same as the coefficient of x4 in the expansion of
(1+x+x2+2x3+x4)n
= ((1+x)+x2(1+x)+x3(1+x))n
= (1+x)n (1+x2+x3)n
Now can someone finish it?
4Neither my ans nor Anant Sir;s ans satisfy the cases of n=1 & n=2
Then what's the solution & where's the mistake ????????
62The mistake is that your factorisation is not right...
it is x5 not x3
Btw what you have done is really what kills this question....
Just see it more closely [1]
4[(1+x) + x2(1+x) + x3(1+x)+x5(1+x) ]n
= (1+x)n(1+x2+x3+x5)n
= (1+x)n (1+x2)n (1+x3)n
coeff = nC0 nC0 nC2 + nC1 nC0nC1 + nC2 nC1nC0
= 1/2 [n3 + 2n2 - n ]
VERIFICATION :
If I take n=1
coeff of x4 = 1
If I take n=2
coeff of x4= 7
62Sorry Anant sir.. I am stealing your thunder :D
Is it obvious that the coeff of x4 in (1 + x + x2 + 2x3 + x4 + x5 + x6)n is same as the highest power of 4 in (1 + x + x2 + 2x3 + x4)n because x^5, x^6 when multiplied to any other term will give a power of greater than or equal to 4....
Now can you work some more on this one? GIve some hints so taht we can move forward...
What do you think you can do from here on?
24I dont have the answer....can u tell me the steps ,how u did it ??