[tex]\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}[/tex]
The rows add up to [tex]1,2,4,8,16[/tex], respectively. (Notice they're all powers of 2)
The sum of the numbers in row [tex]n[/tex] is [tex]2^{n-1}[/tex].
The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When [tex]n=1[/tex],
[tex](1+x)^1=1+x=\dbinom10+\dbinom11x[/tex]
so the base case holds. Assume the claim holds for [tex]n=k[/tex], so that
[tex](1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k[/tex]
Use this to show that it holds for [tex]n=k+1[/tex].
[tex](1+x)^{k+1}=(1+x)(1+x)^k[/tex]
[tex](1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)[/tex]
[tex](1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}[/tex]
Notice that
[tex]\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}[/tex]
[tex]\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}[/tex]
[tex]\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}[/tex]
[tex]\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}[/tex]
[tex]\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}[/tex]
[tex]\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}[/tex]
So you can write the expansion for [tex]n=k+1[/tex] as
[tex](1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}[/tex]
and since [tex]\dbinom{k+1}0=\dbinom{k+1}{k+1}=1[/tex], you have
[tex](1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}[/tex]
and so the claim holds for [tex]n=k+1[/tex], thus proving the claim overall that
[tex](1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n[/tex]
Setting [tex]x=1[/tex] gives
[tex](1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n[/tex]
which agrees with the result obtained for part (c).
