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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.

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    $\begingroup$ How about letting $a = 1$ and $b = x$ and then $x = 1$? $\endgroup$ – k.stm Aug 20 '16 at 5:59
  • $\begingroup$ In which case $(1+1)^n = 2^n$... Okay, my mind's combusting. @k.stm $\endgroup$ – Mad Banners Aug 20 '16 at 6:03
  • $\begingroup$ Notice if the coefficients of $(1+x)^n $ are a,b,c,d,e,f,g.... then the coefficients of $(1+x)^n (1+x) $ are a,a+b,b+c,c+d, etc. Which is basically the same way you construct pascal triangle. Ex. $(1+x)^2= 1+2x+x^2$. So $(1+x)^3 = (1 + 2x + x^2)(1+x)= (1 + 2x + x^2) + (x+2x^2 +x^3) = 1 + (1+2)x + (2+1)x^2+x^3=1 +3x +3x^2+x^3$. See how that works? $\endgroup$ – fleablood Aug 20 '16 at 6:09
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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 $\binom {n}{k}$?

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  • $\begingroup$ Ahhh. k being the $(r+1)$th term of the expansion. What's the point of finding the 1st term using the general term formula though? Why not just $a^n$? $\endgroup$ – Mad Banners Aug 20 '16 at 6:09
  • $\begingroup$ @MadBanners Not sure I understand what you're referring to.. $n$ refers to the $n+1$-th row and $k$ refers to the element in the $n+1$-th row. For example the third row is $1 \ \ 2 \ \ 1$ which correspond to $\binom 20, \binom 21, \binom 22$ $\endgroup$ – Ant Aug 20 '16 at 6:26
  • $\begingroup$ Fair enough. What I mean is that the $(r+1)th$ — aka general — term in $(a+b)^n$ is $T_r+1=\binom{n}{r}a^{n-r}b^r$. For example, the first term in $(1+x)^n$ is $\binom{n}{0}a^{n-0}b^0 = \binom{n}{0}.$ $\endgroup$ – Mad Banners Aug 20 '16 at 6:39
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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 $(1+x)^n = 2^n = \binom{n}{0}+\binom{n}{1}+\binom{n}{2}+....+\binom{n}{n-1}+\binom{n}{n}$

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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 $$ a_0 + (a_0 + a_1) + (a_1 + a_2) + \cdots + (a_{k-1} + a_k) + a_k\\ = 2a_0 + 2a_1 + \cdots + 2a_{k-1} + 2a_k \\ = 2(a_0 + a_1 + \cdots + a_k) = 2\cdot 2^k = 2^{k+1} $$

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