Show that $(1+x)^n$ = .... Hence deduce that .... = $2^n$

Question

Full question:

Show that $(1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+....+\binom{n}{n-1}x^{n-1}+\binom{n}{n}x^n$.

Hence deduce that $\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+....+\binom{n}{n-1}+\binom{n}{n}=2^n$.

The preceding subquestions asked me to write down the first 5 rows of Pascal's triangle, to find the sum of the numbers in rows 1, 2, 3, 4 and 5, and to copy and complete: "The sum of the numbers in row n of Pascal's triangle is......"

I wrote down the first 5 rows of Pascal's triangle, in the same form as the triangle. Obviously the very first row is the zeroth row, so that had to be taken into account.

The sum of the numbers in said rows was $2$ (row 1), $4$ (row 2), $8$ (row 3), $16$ (row 4) and $32$ (row 5), leading me to state that the sum of the numbers in row n of Pascal's triangle is equal to $2^n$.

What I think may be useful here is the formula

$(a+b)^n = a^n + \binom{n}{1}a^{n-1}b +....+\binom{n}{r}a^{n-r}b^r+....+b^n$ where $\binom{n}{r}$ is the binomial coefficient of $a^{n-r}b^r$ and $r = 0, 1, 2, 3,....,n.$

I can kind of see how this relates, but not quite.

Answer 1

Show that $(1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+....+\binom{n}{n-1}x^{n-1}+\binom{n}{n}x^n$.

This can be done with induction, I will show the inductive step only

$(1+x)^{n+1} = (1+x)(1+x)^n = (1+x)(\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+....+\binom{n}{n-1}x^{n-1}+\binom{n}{n}x^n)$

$=\binom{n}{0} + \binom{n}{n}x^{n+1} + \sum_{k=1}^n x^k(\binom{n}{k} + \binom{n}{k-1}) $

$\binom{n}{k} + \binom{n}{k-1} = \frac{n!}{k!(n-k)!} + \frac{n!}{(k-1)! (n-k+1)!} = n!(\frac{n-k+1}{k!(n-k+1)!} + \frac{k}{k!(n-k+1)!})$

$=n! \frac{n+1}{k!(n-k+1)!} = \binom{n+1}{k}$ which is the coefficient of $x^k$ in the $\sum$ above.

i.e. $(1+x)^{n+1} = 1 + x^{n+1} + \sum_{k=1}^{n}x^k \binom{n+1}{k}$

Setting $x=1$ in $(1+x)^n = 2^n = \binom{n}{0}+\binom{n}{1}+\binom{n}{2}+....+\binom{n}{n-1}+\binom{n}{n}$

Answer 2

Take $a=x $ and $b=1 $ to get your formula. To get the second formula choose also $ x = 1$ It's pretty clear though that you can't use the binomial expansion formula to "prove" yours as they are basically the same.

For the hint, do you know that the coefficients in the pascal triangle are the $\binom {n}{k}$?

Answer 3

Here is a proof of the hint using induction. You've already checked the first six cases, so I'll go straight to the induction step.

Let $a_0, \ldots, a_k$ be the elements of row $k$, and assume $a_0+\cdots +a_k = 2^k$. Now, what are the elements of row $k+1$ expressed in the $a_i$? Well, the first one is just $a_0$ (that's the $1$ on the edge). After that it goes $a_0+a_1, a_1+a_2$, and so on, until the last one which is just $a_k$ (the $1$ on the other edge).

We want the sum of all these terms. Writing it down, we get $$ a_0 + (a_0 + a_1) + (a_1 + a_2) + \cdots + (a_{k-1} + a_k) + a_k\\ = 2a_0 + 2a_1 + \cdots + 2a_{k-1} + 2a_k \\ = 2(a_0 + a_1 + \cdots + a_k) = 2\cdot 2^k = 2^{k+1} $$