Question

The coefficient of x^7 in the expansion of (1-x-x^2 +x^3)^6 is?

Answer 1

pls explain using multinomial theorem..

Answer 2

@imqwerty

Answer 3

multinomial theorem says that in \(\left(a_1+a_2+a_3+....+a_m\right)^n\)
the coefficient of \(a_1^{p_1}a_2^{p_2}...a_m^{p_m}\) =\(\large \frac{n!}{p_1!.p_2!...p_m!}\)
where \(p_1+p_2+...p_m=n\)

1st we need to know the ways to get \(x^7\)
and then find the coefficient of each case and then add them all

Answer 4

ok..let me try..

Answer 5

okay

Answer 6

there 7 such cases right?

Answer 7

@imqwerty

Answer 8

um i didn't do the calculation tho

Answer 9

and sorry it took so long..

Answer 10

its okay
ima do the calculations

Answer 11

there were so many possibilities...(trial and error)..

Answer 12

yes
its better to convert such cases into a binomial theorem applicable form

Answer 13

how can we convert this?

Answer 14

we can write it like this-

\(1-x-x^2+x^3=(x-1)^3+2(x-1)^2\)

Answer 15

shouldn't it be 2(x+1)^2 over there?

Answer 16

oh no..

Answer 17

but still.. its not correct right?

Answer 18

its coming alright tho?
\((x-1)^3+2(x-1)^2 = x^3-1-3x^2+3x +2x^2+2-4x\)
=\(x^3-x^2-x+1\)

Answer 19

yeah its correct! sry..

Answer 20

its okay =]
after this conversion we can apply binomial
and cases reduce in this method

Answer 21

yes..

Answer 22

Thanks!

Answer 23

np :)