\(\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}\)
The rows add up to \(1,2,4,8,16\), respectively. (Notice they're all powers of 2)
The sum of the numbers in row \(n\) is \(2^{n-1}\).
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 \(n=1\),
\((1+x)^1=1+x=\dbinom10+\dbinom11x\)
so the base case holds. Assume the claim holds for \(n=k\), so that
\((1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\)
Use this to show that it holds for \(n=k+1\).
\((1+x)^{k+1}=(1+x)(1+x)^k\)
\((1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)\)
\((1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}\)
Notice that
\(\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}\)
\(\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}\)
\(\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}\)
\(\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}\)
\(\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}\)
\(\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}\)
So you can write the expansion for \(n=k+1\) as
\((1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}\)
and since \(\dbinom{k+1}0=\dbinom{k+1}{k+1}=1\), you have
\((1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}\)
and so the claim holds for \(n=k+1\), thus proving the claim overall that
\((1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n\)
Setting \(x=1\) gives
\((1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n\)
which agrees with the result obtained for part (c).
