Proof.

Step 1:

For \(n=1\) the statement reads \((a+b)^{1}=a+b\), which is true. (The case \(n=0\) also holds, both sides being \(1\).)

Step 2:

We assume that it is true for \(n=k\) \[(a+b)^k=a^k+ \binom {k}{1} a^{k-1}b+\cdots + \binom {k}{r} a^{k-r}b^r+\cdots +b^k \]

Step 3:

We show it then holds for \(n=k+1\) \[(a+b)^{k+1}=a^{k+1}+ \binom {k+1}{1} a^{(k+1)-1}b+\cdots + \binom {k+1}{r} a^{(k+1)-r}b^r+\cdots +b^{k+1} \] Take step 2. \[(a+b)^k=a^k+ \binom {k}{1} a^{k-1}b+\cdots + \binom {k}{r} a^{k-r}b^r+\cdots +b^k \] \(\therefore \) The \(r^{\text {th}}\) term for \(n=k\)

\begin {align*} (a+b)^k & = a^k+ka^{k-1}b+\cdots +\frac {k(k-1)(k-2)\cdots (k-r+2)}{(r-1)!}a^{k-r+1}b^{r-1}+\\ & \cdots +b^k\quad \cdots (2)\, \text {True}\\\\ (a+b)\Big \{(a+b)^k &=a^k+ \binom {k}{1} a^{k-1}b+\cdots + \binom {k}{r} a^{k-r}b^r+\cdots +b^k\Big \}\\\\ (a+b)^{k+1} &=a^{k+1}+ka^kb+\cdots +\frac {k(k-1)(k-r+2)a^{k-r+2}b^{r-1}}{(r-1)!}+\cdots +ab^k\\ &+ \cdots +\frac {k(k-1)\cdots (k-1+3}{(r-2)!}a^{k-r+3}b^{r+2}+\cdots +kab^k+b^{r+1} \end {align*}

The terms prove that through \(ab^k\) represents the right number of (2) multiplied by \(a\). To obtain the \(r^{\text {th}}\) term in the result, we have written the \(r^{\text {th}}\) term in the first half of right member and the \((r-1)^{\text {th}}\) term which is equal to the second, this both contains \(a^{k-r+2}b^{r-1}\).

\[\frac {k(k-1)\cdots (k-r+2)}{(r-1)!}+\frac {k(k-1)\cdots (k-r+3)}{(r-2)!}\] \[=\frac {k(k-1)\cdots (k-r+3)(k-r+2)+k(k-1)\cdots (k-r+3)}{(r-1)!(r-2)!}\]

Independent \(x\).
□

Solution.

Take the general form \( \binom {n}{r} a^{n-r}b^r \), \(a=3x\) and \(b=-\frac {1}{x}\)

\begin {align*} \implies \binom {n}{r} (3x)^{n-r}\left (-\frac {1}{x}\right )^r & = \binom {n}{r} 3^{n-r}\cdot x^{n-r}(-1)^r x^{-r}\\ & = \binom {n}{r} 3^{n-r}(-1)^r x^{n-2r} \end {align*}

\(n=20\)

\(\implies \binom {20}{r} 3^{20-r}(-1)^rx^{20-2r} \)

Solve, \(20-2r=0\quad \implies 2r=20\quad \implies r=10\)

\[n=20,\quad r=10,\quad a=3x,\quad b=-\frac {1}{x}\]

\[\therefore \quad \binom {20}{10} (3)^{10}(-1)^{10} \]

\[ \binom {20}{10} =\frac {20!}{10!10!} \] So, the 11\(^{\text {th}}\) term is independent of \(x\).

When expanding expressions of the form \((a+b)^n\) where \(n\) is negative or rational number.

In this case we use the formula: \[(1+x)^n=1 + nx +\frac {n(n-1)x^2}{2!}+\frac {n(n-1)(n-2)x^3}{3!}+\cdots +\cdots \] This infinite series is valid in the region \(|x|<1\).

Solution.

\begin {align*} \frac {1}{\sqrt {3+2x}} & = (3+2x)^{-1/2}\\ & = \left [3\left (1+\frac {2x}{3}\right )\right ]^{-1/2},\quad n=-\frac {1}{2},\, x=\frac {2x}{3}\\\\ & = 3^{-1/2}\left (1+\frac {2x}{3}\right )^{-1/2}\\ & = \frac {1}{\sqrt {3}}\left [1+\left (-\frac {1}{2}\right ) \left (\frac {2x}{3}\right )+\frac {\left (\frac {-1}{2}\right ) \left (\frac {-3}{2}\right ) \left (\frac {2x}{3}\right )^2}{2!}+\frac {\left (\frac {-1}{2}\right )\left (\frac {-3}{2}\right )\left (\frac {-5}{2}\right )\left (\frac {2x}{3}\right )^3}{3!}+\cdots \right ]\\\\ & = \frac {1}{\sqrt {3}}\left [1-\frac {1}{3}x+\frac {12x^2}{72}-\frac {15(8)x^3}{(8)(27)6}+\cdots \right ]\\\\ & = \frac {1}{\sqrt {3}}\left [1-\frac {1}{3}x+\frac {x^2}{6}-\frac {5x^3}{54}+\cdots \right ]\\ \end {align*}

This is valid for: \(|x|<1\)

i.e \(\bigg |\frac {2}{3}x\bigg |<1\implies |x|<\frac {3}{2}\)