Question

$\left(\begin{array}{c}5\\ 0\end{array}\right){\left(\frac{1}{4}\right)}^5+\left(\begin{array}{c}5\\ 1\end{array}\right){\left(\frac{1}{4}\right)}^4\left(\frac{3}{4}\right)+\left(\begin{array}{c}5\\ 2\end{array}\right){\left(\frac{1}{4}\right)}^3{\left(\frac{3}{4}\right)}^2+\left(\begin{array}{c}5\\ 3\end{array}\right){\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^3+\left(\begin{array}{c}5\\ 4\end{array}\right){\left(\frac{1}{4}\right)}^1{\left(\frac{3}{4}\right)}^4+\left(\begin{array}{c}5\\ 5\end{array}\right){\left(\frac{3}{4}\right)}^5=?$

Short answer

The value of$\left(\begin{array}{c}5\\ 0\end{array}\right){\left(\frac{1}{4}\right)}^5+\left(\begin{array}{c}5\\ 1\end{array}\right){\left(\frac{1}{4}\right)}^4\left(\frac{3}{4}\right)+\left(\begin{array}{c}5\\ 2\end{array}\right){\left(\frac{1}{4}\right)}^3{\left(\frac{3}{4}\right)}^2+\left(\begin{array}{c}5\\ 3\end{array}\right){\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^3+\left(\begin{array}{c}5\\ 4\end{array}\right){\left(\frac{1}{4}\right)}^1{\left(\frac{3}{4}\right)}^4+\left(\begin{array}{c}5\\ 5\end{array}\right){\left(\frac{3}{4}\right)}^5=1$

Step-by-step solution

Step 1. Given information

Here

$\left(\begin{array}{c}5\\ 0\end{array}\right){\left(\frac{1}{4}\right)}^5+\left(\begin{array}{c}5\\ 1\end{array}\right){\left(\frac{1}{4}\right)}^4\left(\frac{3}{4}\right)+\left(\begin{array}{c}5\\ 2\end{array}\right){\left(\frac{1}{4}\right)}^3{\left(\frac{3}{4}\right)}^2+\left(\begin{array}{c}5\\ 3\end{array}\right){\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^3+\left(\begin{array}{c}5\\ 4\end{array}\right){\left(\frac{1}{4}\right)}^1{\left(\frac{3}{4}\right)}^4+\left(\begin{array}{c}5\\ 5\end{array}\right){\left(\frac{3}{4}\right)}^{5}isgiven$

we have to find out the value of the given statement

Step  2. Description of finding the value of  given question

By using the binomial theorem we can find the value for this problem.

Binomial Theorem states that

$$\begin{gathered} Letxandaberealnumbers.Foranypositiveintegern,wehave \\ {(x+a)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+\left(\begin{array}{c}n\\ 2\end{array}\right)a^2{x}^{n-2}+.....+\left(\begin{array}{c}n\\ n\end{array}\right)a^n \\ Thegivenquesion\left(\begin{array}{c}5\\ 0\end{array}\right){\left(\frac{1}{4}\right)}^5+\left(\begin{array}{c}5\\ 1\end{array}\right){\left(\frac{1}{4}\right)}^4\left(\frac{3}{4}\right)+\left(\begin{array}{c}5\\ 2\end{array}\right){\left(\frac{1}{4}\right)}^3{\left(\frac{3}{4}\right)}^2+\left(\begin{array}{c}5\\ 3\end{array}\right){\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^3+\left(\begin{array}{c}5\\ 4\end{array}\right){\left(\frac{1}{4}\right)}^1{\left(\frac{3}{4}\right)}^4+\left(\begin{array}{c}5\\ 5\end{array}\right){\left(\frac{3}{4}\right)}^{5}isgiven \\ isintheformofbinomialtheoremexpansion.Socomparedtoexpandedform \\ wegetn=5,x=\left(\frac{1}{4}\right)anda=\left(\frac{3}{4}\right)thenbyapplyingbinomialexpansionformweget \\ {\left(\frac{1}{4}+\frac{3}{4}\right)}^5={\left(\frac{4}{4}\right)}^5 \\ ={\left(1\right)}^5=1,sovalueof\left(\begin{array}{c}5\\ 0\end{array}\right){\left(\frac{1}{4}\right)}^5+\left(\begin{array}{c}5\\ 1\end{array}\right){\left(\frac{1}{4}\right)}^4\left(\frac{3}{4}\right)+\left(\begin{array}{c}5\\ 2\end{array}\right){\left(\frac{1}{4}\right)}^3{\left(\frac{3}{4}\right)}^2+\left(\begin{array}{c}5\\ 3\end{array}\right){\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^3+\left(\begin{array}{c}5\\ 4\end{array}\right){\left(\frac{1}{4}\right)}^1{\left(\frac{3}{4}\right)}^4+\left(\begin{array}{c}5\\ 5\end{array}\right){\left(\frac{3}{4}\right)}^{5}is1 \end{gathered}$$