Question

What is the coefficient `a_6` in `(1+x)^21+cdots+(1+x)^30` ?

Answer 1

`""_21C_6+""_22C_6+...+""_30C_6.`

Explanation:

Recall that, `(1+x)^n=sum_(r=0)^(r=n)""_nC_rx^r.`

Therefore, the co-eff. of `x^r" in the expansion of "(1+x)^n` is `""_nC_r.`

Hence, the co-effs. of `x^6` in `(1+x)^n, where, n=21,22,...,30,`

are, `""_21C_6, ""_22C_6, ..., ""_30C_6,` resp.

`:.`The Reqd. Co-eff.=`""_21C_6+""_22C_6+...+""_30C_6.`

Answer 2

See below.

Explanation:

`(1+x)^21+cdots+(1+x)^30=(1+x)^21(1+(1+x)+cdots+(1+x)^9)=`

`=(1+x)^21((1+x)^10-1)/((1+x)-1) = 1/x((1+x)^31-(1+x)^21)`

Now

`(1+x)^n=sum_(k=1)^n ((k),(n))x^k` then

`a_6 = ((7),(31))-((7),(21)) = (31!)/(7! (31 - 7)!) - (21!)/(7! (21 - 7)!) = 2513295`

Answer 3

`Coef(x^6) = 2513295`

Explanation:

We seek the coefficient of `x^6` in:

`S = (1+x)^21 + (1+x)^22 + ... + (1+x)^30`

Consider the sum

`S_n = (1+x)^0 + (1+x)^1 + (1+x)^2 + ... + (1+x)^n`

These term for a GP with `(n+1)` terms:

`a = 1`
`r = 1+x`

And so we can use the GP sum formula to evaluate the sum of the first `n` terms:

`S = a(1-r^n)/(1-r)`
`\ \ = (1-(1+x)^n)/(1-(1+x))`
`\ \ = ((1+x)^n-1)/x`

Then we can find `S` using:

`S = S_31 - S_21`

`\ \ = ((1+x)^31-1)/x - ((1+x)^21-1)/x`

`\ \ = (1+x)^31/x- (1+x)^21/x`

We can determine the coefficient of `x^r` in a binomial expansion `(1+x)^n` using the combination:

`""_nC^r = ( (n), (r) ) = (n!)/((n-r)!r!)`

We can evaluate the combination directly using factorials, or more likely using a calculators in-built function for combinations.

As dividing by `x` will reduce the power of `x` in the binomial expansion then the coefficient of `x^6` that we seek is given by

`Coef(x^6) = Coef(x^7){(1+x)^31} - Coef(x^7){(1+x)^21}`
`" " = ( (31), (7) ) - ( (21), (7) )`
`" " = 26295755 - 116280`
`" " = 2513295`