55. Apply the binomial theorem of Newton to find the value of the following sums:

a) $C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+\ldots+C_{n}^{n}$

b) $C_{n}^{0}-C_{n}^{1}+C_{n}^{2}-\ldots+(-1)^{n} C_{n}^{n}$

c) $C_{n}^{0}+\frac{1}{2} C_{n}^{1}+\frac{1}{3} C_{n}^{2}+\ldots+\frac{1}{n+1} C_{n}^{n}$

d) $C_{n}^{1}+2 C_{n}^{2}+3 C_{n}^{3}+\ldots+n C_{n}^{n}$,

e) $C_{n}^{1}-2 C_{n}^{2}+3 C_{n}^{3}-\cdots+(-1)^{n-1} n C_{n}^{n}$,

f) $C_{n}^{0}-C_{n}^{1}+C_{n}^{2}-\ldots+(-1)^{m} C_{n}^{m}$

g) $C_{n}^{k}+C_{n+1}^{k}+C_{n+2}^{k}+\ldots+C_{n+m}^{k}$

h) $C_{2 n}^{0}-C_{2 n-1}^{1}+C_{2 n-2}^{2}-\ldots+(-1)^{n} C_{n}^{n}$

i) $C_{2 n}^{n}+2 C_{2 n-1}^{n}+4 C_{2 n-2}^{n}+\ldots+2^{n} C_{n}^{n}$

j) $\left(C_{n}^{0}\right)^{2}+\left(C_{n}^{1}\right)^{2}+\left(C_{n}^{2}\right)^{2}+\ldots+\left(C_{n}^{n}\right)^{2}$,

k) $\left(C_{n}^{0}\right)^{2}-\left(C_{n}^{1}\right)^{2}+\left(C_{n}^{2}\right)^{2}-\ldots+(-1)^{n}\left(C_{n}^{n}\right)^{2}$

l) $C_{n}^{0} C_{m}^{k}+C_{n}^{1} C_{m}^{k-1}+C_{n}^{2} C_{m}^{k-2}+\ldots+C_{n}^{l} C_{m}^{0}$

Some of the sums included in this problem will be encountered below in a different context, sometimes in a more general form. For example, the sum in problem m) will be calculated by a different method in the solutions of problems 58a), 99v) (70b). The result of problem d) will be significantly generalized in the solution of problem 79v). The sum in problem i) will be determined by a different method in
55. a) $C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+\ldots+C_{n}^{n}=(1+1)^{n}=2^{n}$.

b) $C_{n}^{0}-C_{n}^{1}+C_{n}^{2}-\ldots+(-1)^{n} C_{n}^{n}=(1-1)^{n}=0$.

c) Using the fact that

$$
\frac{n+1}{k+1} C_{n}^{k}=\frac{(n+1) n(n-1) \ldots(n-k+1)}{1 \cdot 2 \cdot 3 \ldots k(k+1)}=C_{n}^{k+1}
$$

Therefore,

$$
\begin{aligned}
& (n+1)\left\{C_{n}^{0}+\frac{1}{2} C_{n}^{1}+\frac{1}{3} C_{n}^{2}+\ldots+\frac{1}{n+1} C_{n}^{n}\right\}= \\
& =C_{n+1}^{1}+C_{n+1}^{2}+\ldots+C_{n+1}^{n+1}=(1+1)^{n+1}-1=2^{n+1}-1
\end{aligned}
$$

and

$$
C_{n}^{0}+\frac{1}{2} C_{n}^{1}+\frac{1}{3} C_{n}^{2}+\ldots+\frac{1}{n+1} C_{n}^{n}=\frac{2^{n+1}-1}{n+1}
$$

d) Using the fact that

$$
k C_{n}^{k}=n \frac{(n-1)(n-2) \ldots(n-k+1)}{1.2 \ldots(k-1)}=n C_{n-1}^{k-1}
$$

Therefore,

$$
\begin{aligned}
C_{n}^{1}+2 C_{n}^{2} & +3 C_{n}^{3}+\ldots+n C_{n}^{n}= \\
& =n\left(C_{n-1}^{0}+C_{n-1}^{1}+C_{n-1}^{2}+\ldots+C_{n-1}^{n-1}\right)= \\
& =n(1+1)^{n-1}=2^{n-1} n_{.}
\end{aligned}
$$

e) Similarly to the solution of problem d), we get:

$$
\begin{aligned}
& C_{n}^{1}-2 C_{n}^{2}+3 C_{n}^{3}-\ldots+(-1)^{n-1} n C_{n}^{n}= \\
& \quad=n\left[C_{n-1}^{0}-C_{n-1}^{1}+C_{n-1}^{2}-\ldots+(-1)^{n-1} C_{n-1}^{n-1}\right]= \\
& \quad=n(1-1)^{n-1}=0 .
\end{aligned}
$$

f) The desired sum is equal to the coefficient of $x^{n}$ in the following polynomial:

$$
x^{n}(1-x)^{n}+x^{n-1}(1-x)^{n}+\ldots+x^{n-m}(1-x)^{n}
$$

Transforming this polynomial, we get:

$$
\begin{aligned}
& x^{n}(1-x)^{n}+x^{n-1}(1-\dot{x})^{n}+ \\
& +x^{n-2}(1-x)^{n}+\ldots+x^{n-m}(1-x)^{n}= \\
& =(1-x)^{n}\left(x^{n}-1-x^{n-1}+x^{n-2}+\ldots+x^{n-m}\right)= \\
& =(1-x)^{n} \frac{x^{n+1}-x^{n-m}}{x-1}=-(1-x)^{n-1}\left\{x^{n+1}-x^{n-m}\right\}= \\
& =x^{n-m}(1-x)^{n-1}-x^{n+1}(1-x)^{n} .
\end{aligned}
$$

Therefore, the desired coefficient of $x^{n}$ is $(-1)^{m} C_{n-1}^{m}$ for $m \leqslant n-1$ and is 0 for $m=n$ (compare with the result of problem b)).

g) The desired sum is equal to the coefficient of $x^{k}$ in the