Simplify: $2^k~ { }^n C_0 { }^n C_k - 2^{k-1} ~{ }^n C_1 { }^{n-1} C_{k-1} + 2^{k-1} { }^n C_2 { }^{n-2} C_k … (-1)^k { }^n C_k { }^{n-k} C_0$

Solution:

$i^{th}$ term $\rightarrow \sum_{i=0}^{i-k} (-1)^i 2^{k-i} { }^n C_i { }^{n-i}C_{k-i}$

$\frac{k !}{k !}$ $\cdot \frac{n !}{i ! (n-i)!}$ $\cdot \frac{(n-i)!}{(k-i)!(n-k)}$

$\sum_{i=0}^k (-1)^i 2^{k-i} ~ { }^n C_k { }^k C_i$

$={ }^n C_k [ \sum_{i=0}^{k} (-1)^i 2^{k-i} { }^kC_i] = { }^n C_k$

${ }^k C_0 \cdot 2^k (-1)^0 + { }^k C_1 2^{k-1} (-1)^1 + { }^k C_2 2^{k-2} (-1)^2 + …[2+(-1)]^k$

${ }^n C_r \quad\left(\begin{array}{l}n \\ r\end{array}\right)$

$S_k = 3^k ( ^{100} _{ 0 }) ( ^{100} _{ k }) -3^{k-1} ( ^{100} _{ 1 }) ( ^{99} _{ k-1 }) + 3^{k-2} ( ^{100} _{ 2 }) ( ^{98} _{ k-2 }) - … + (-1)^k ( ^{100} _{ k }) ( ^{100} _{ 0 })$

$V_k = (1/2)^k S_k = M(100,k)$

A) $\sum_{k=0}^{100} S_k S-{100-k}$

B) $M(100,49)+M(100,50)$

$S_k =3^k { }^{100} C_0 { }^{100} C_k - 3^{k-1} { }^{100} C_1 { }^{99} C_{k-1} +…(-1)^k { }^{100} C_k { }^{100-k} C_0$

$= \sum_{i=0}^k 3^{k-i} \cdot (-1)^i \cdot { }^{100} C_i \cdot { }^{100-i} C_{k-i}$

$\frac{k!}{k!} \cdot \frac{100!}{i!(100-i)!} \cdot \frac{(100-i)!}{(k-i)!(100-k)!}$

$=\sum_{i=0}^k 3^{k-i} (-1)^i { }^{100} C_k \cdot { }^k C_i = { }^{100} C_k \cdot \sum_{i=0}^k 3^{k-i} \cdot (-1)^i { }^k C_i$

$(3-1)(3-1)….(3-1)$

$(3-1)(3-1)….(3-1) \rightarrow k$

$(3-1)(3-1).. \rightarrow k-i$

$…(3-1)\rightarrow i$

${ }^{100} C_k \cdot \sum_{i=0}^k 3^{k-i} \cdot (-1)^i { }^k C_i \rightarrow (3-1)^k$
$S_k={ }^{100} C_k \cdot 2^k$

$S_{100-k}={ }^{100} C_{100-k} \cdot 2^{100-k}$

$\sum_{k=0}^{100} S_k S_{100-k}=\sum_{k=0}^{100}{ }^{100} C_k \cdot{ }^{100} C_{100-k} \cdot 2^{100}$

$=2^{100} \sum_{k=0}^{100}{ }^{100} c_k{ }^{100} C_{100-k}$

$(x+1 / x)^{200}=(x+1 / x)^{100}(x+1 / x)^{100}$

${ }^{200} C_{100}=C_0 \cdot c_{100}+c_1 \cdot c_{99}+c_2 \cdot c_{98}+\cdot \cdot$

$={ }^{200} C_{100} \cdot 2^{100}$

$M(100, k)=(\frac{1}{2})^k S_k$

$=(1 / 2)^k \cdot{ }^{100} C_k \cdot 2^k={ }^{100} C_k$

$M(100,49)+M(100,50)={ }^{100} C_{49}+{ }^{100} C_{50}$

${ }^{101} C_{50}={ }^{101} C_{51}$

$1 + x, 1 + x + x^2, 1 + x + x^2 + x^3, …. 1 + x + x^2 + … x^n$

$(1 + x)(1 + x + x^2)(1 + x + x^2 + x^3)(…) … (1 + x + x^2 + … + x^n)$

$= a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ….. a_N x^N$

1. How many terms in the expansion?

2. Show that coeffs equi-distant are equal from beginning and end.

3. Sum of odd coeffs = Sum of even coeffs

$=\frac{(n+1)!}{2}$

$x^{1+2+3……n}= x^{\frac{n(n+1)}{2}}$

${\frac{n(n+1)}{2}}+1$

$(1+x)(1+x+x^2)(…)…(1+x+x^2 + x+…x^n=a_0 +a_1x + … a_N x^N$

$(1+x^{-1})(1+x^{-1} + x^{-2})…(1+x^{-1} +x^{-2} + … x^{-n})=a_0 + a_1 x^{-1}+…a_Nx^{-N}$

$x^{-N}(1+x)(1+x+x^2)…(x^n + x^{n-1} + … +x^2 +x +1)=x^{-N}(a_0 + a_1x + … a_N x^N)$

$0=a_0-a_1+a_2-a_3+… \Rightarrow$ Sum of odd terms

= Sum of even terms

$2 \times 3 \times 4 \times … (n+1) = 2 \times$ Sum of odd