Question

Question: Prove that for any positive integer n, ${\mathbf{2}}^{\mathbf{n}}\mathbf{=}\left[\begin{array}{c}n\\ 0\end{array}\right]\mathbf{+}\left[\begin{array}{c}n\\ 1\end{array}\right]\mathbf{+}\left[\begin{array}{c}n\\ 2\end{array}\right]\mathbf{+}\left[\begin{array}{c}n\\ 3\end{array}\right]\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left[\begin{array}{c}n\\ n\end{array}\right]$.

[Hint]

Short answer

Answer:

It is proved that$2^{\mathrm{n}}=\left[\begin{array}{c}\mathrm{n}\\ 0\end{array}\right]+\left[\begin{array}{c}\mathrm{n}\\ 1\end{array}\right]+\left[\begin{array}{c}\mathrm{n}\\ 2\end{array}\right]+\left[\begin{array}{c}\mathrm{n}\\ 3\end{array}\right]+.............\left[\begin{array}{c}\mathrm{n}\\ \mathrm{n}\end{array}\right]$

Step-by-step solution

Definition of the Binomial Theorem

This theorem states that let R be a commutative rind and . Then each positive integer n,

${\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}}^{\mathbf{n}}\mathbf{=}{\mathit{a}}^{\mathbf{n}}\mathbf{+}\mathbf{\left[}\begin{array}{c}\mathbf{n}\\ \mathbf{1}\end{array}\mathbf{\right]}{\mathit{a}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathit{b}\mathbf{+}\mathbf{\left[}\begin{array}{c}\mathbf{n}\\ \mathbf{2}\end{array}\mathbf{\right]}{\mathit{a}}^{\mathbf{n}\mathbf{-}\mathbf{2}}{\mathit{b}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{\left[}\begin{array}{c}\mathbf{n}\\ \mathbf{n}\mathbf{-}\mathbf{1}\end{array}\mathbf{\right]}\mathbf{}\mathit{a}{\mathit{b}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{+}{\mathit{b}}^{\mathbf{n}}$

Prove that for any positive integer n,2n=[n0]+[n1]+[n2]+[n3]+.............[nn]

Consider the expansion as follows.

$2^n$

This expansion can be written as follows.

$2^n={\left(1+1\right)}^n$

Let a = 1 and b = 1

Apply the binomial theorem.

$$\begin{gathered} {\left(1+1\right)}^n=\left[\begin{array}{c}n\\ 0\end{array}\right]1^{n-0}{\left(1\right)}^0+\left[\begin{array}{c}n\\ 1\end{array}\right]1^{n-1}{\left(1\right)}^1+\left[\begin{array}{c}n\\ 2\end{array}\right]1^{n-2}{\left(1\right)}^2+\left[\begin{array}{c}n\\ 3\end{array}\right]1^{n-3}{\left(1\right)}^3+...+\left[\begin{array}{c}n\\ n\end{array}\right]1^{n-n}{\left(1\right)}^n \\ =\left[\begin{array}{c}n\\ 0\end{array}\right]1^{n-0}+\left[\begin{array}{c}n\\ 1\end{array}\right]1^{n-1}+\left[\begin{array}{c}n\\ 2\end{array}\right]1^{n-2}+\left[\begin{array}{c}n\\ 3\end{array}\right]1^{n-3}+...+\left[\begin{array}{c}n\\ n\end{array}\right]1^{n-n} \\ =\left[\begin{array}{c}n\\ 0\end{array}\right]+\left[\begin{array}{c}n\\ 1\end{array}\right]+\left[\begin{array}{c}n\\ 2\end{array}\right]+\left[\begin{array}{c}n\\ 3\end{array}\right]+...+\left[\begin{array}{c}n\\ n\end{array}\right] \end{gathered}$$

From the above expansion, to the power of ant positive integers n is 1 .

Therefore, the expansion of is $2^{\mathrm{n}}=\left[\begin{array}{c}\mathrm{n}\\ 0\end{array}\right]+\left[\begin{array}{c}\mathrm{n}\\ 1\end{array}\right]+\left[\begin{array}{c}\mathrm{n}\\ 2\end{array}\right]+\left[\begin{array}{c}\mathrm{n}\\ 3\end{array}\right]+.............\left[\begin{array}{c}\mathrm{n}\\ \mathrm{n}\end{array}\right]$.