Question

Suppose that bis an integer with $\mathit{b}\mathbf{\ge }\mathbf{7}$. Use the binomial theorem and the appropriate row of Pascal's triangle to find the base- bexpansion of $\left(11\right)\mathbf{}{\mathbf{}}_{\mathbf{b}}^{\mathbf{4}}$[that is, the fourth power of the number ${\left(11\right)}_{\mathbf{b}}$in base-bnotation].

Short answer

The base- expansion of (11)${}_b^4$ is$(14641{)}_b$ .

Step-by-step solution

Use Binomial theorem

Binomial theorem: binomial theorem, statement that for any positive integer n, the nth power of the sum of two numbers x and y may be expressed as the sum of n + 1 terms of the form.

$\mathbf{(}\mathit{x}\mathbf{+}\mathit{y}{\mathbf{)}}^{\mathbf{n}}\mathbf{=}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathbf{ }\left(\begin{array}{l}n\\ j\end{array}\right){\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{j}}{\mathit{y}}^{\mathbf{j}}$

Find ((11)b)4 in base-b  notation

So,$1\cdot b+1\cdot 1=b+1$

$\begin{array}{r}{\left((11{)}_b\right)}^4=(b+1{)}^4\\ =\sum _{i=0}^4 \left(\begin{array}{c}4\\ i\end{array}\right)\cdot b^i\\ =1\cdot 1+4\cdot b+6\cdot b^2+4\cdot b^3+1\cdot b^4\\ =(14641{)}_b\end{array}$

The last step is possible when $b\ge 7$.

Thus, the base- expansion of $\left(11{)}_b^4\text{is}\right(14641{)}_b$.