If $x^n=a_0+a_1(1+x)+\ldots+a_n(1+x)^n=b_0 + b_1(1-x)+\ldots+b_n(1-x)^n$, then for $n=101$, find $(a_{50},b_{50})$

Question

My teacher told me add 1 on both sides of the equation, given $$1+x^n=(a_0+1)+a_1(1+x)+\ldots+a_n(1+x)^n$$

But I don’t see how that’s useful in anyway. Am I misunderstanding the hint? What needs to be done here?

Answer 1

Here is a slightly different hint from your teacher's.

Write $x^n$ in two different ways :

$$x^n = ((1+x)-1)^n = (1-(1-x))^n$$

Expand using binomial theorem and compare with other two appropriate polynomials.

Identify $(a_{50}, b_{50})$.

Answer 2

Here is a method using derivatives

$$x^{n}=\sum_{j=0}^{n}a_{j}(1+x)^{j}$$

Taking derivative $w.r.t$ $x$

$$nx^{n-1}=\sum_{j=1}^{n}a_{j}j(1+x)^{j-1}$$

Again derivate $w.r.t$ $x$.

$$n(n-1)x^{n-2}=\sum_{j=2}^{n}a_{j}j(j-1)(1+x)^{j-2}$$

So, $K^{th}$ derivative will look like this

$$n(n-1)...(n-k+1)x^{n-k}=\sum_{j=k}^{n}a_{j}j(j-1)...(j-k+1)(1+x)^{j-k}$$

$$\implies \frac{n!}{(n-k)!}x^{n-k}=\sum_{j=k}^{n}a_{j}\frac {j!}{(j-k)!}(1+x)^{j-k}$$

Since $n=101$ we have,

$$\frac{101!}{(101-k)!}x^{101-k}=\sum_{j=k}^{101}a_{j}\frac {j!}{(j-k)!}(1+x)^{j-k}$$

Now Choose $k$ wisely so that you will get $a_{50}$

We can find $b_{50}$ using same method.