Answer

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Hint- Here, a special case of binomial theorem will be used.

Since, we have to find the value for ${\left( {99} \right)^5}$ which can be written as \[{\left( {100 - 1} \right)^5}\].

According to Binomial theorem, we know that

\[{\left( {x - 1} \right)^n} = {}^n{C_0}{x^n}{1^0} - {}^n{C_1}{x^{n - 1}}{1^1} + {}^n{C_2}{x^{n - 2}}{1^2} - ..... + {\left( { - 1} \right)^{n - 1}}{}^n{C_{n - 1}}{x^1}{1^{n - 1}} + {\left( { - 1} \right)^n}{}^n{C_n}{x^0}{1^n}\]

where \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] and \[{}^n{C_0} = 1,{}^n{C_1} = n,{}^n{C_{n - 1}} = n,{}^n{C_n} = 1\]

\[ \Rightarrow {\left( {x - 1} \right)^n} = {x^n} - n{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} - ..... + {\left( { - 1} \right)^{n - 1}}nx + {\left( { - 1} \right)^n}\]

In the above equation, put \[x = 100\] and \[n = 5\]

\[ \Rightarrow {\left( {99} \right)^5} = {\left( {100 - 1} \right)^5} = {100^5} - 5 \times {\left( {100} \right)^4} + {}^5{C_2}{\left( {100} \right)^3} - {}^5{C_3}{\left( {100} \right)^2} + 5 \times 100 - 1{\text{ }} \to {\text{(1)}}\]

Now, \[{}^5{C_2} = \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}} = \dfrac{{5!}}{{2!3!}} = \dfrac{{5 \times 4}}{2} = 10\] and \[{}^5{C_3} = \dfrac{{5!}}{{3!\left( {5 - 3} \right)!}} = \dfrac{{5!}}{{3!2!}} = \dfrac{{5 \times 4}}{2} = 10\]

Therefore, equation (1) becomes

\[

\Rightarrow {\left( {99} \right)^5} = {\left( {100 - 1} \right)^5} = {10^{10}} - 5 \times {10^8} + 10 \times {10^6} - 10 \times {10^4} + 500 - 1 \\

\Rightarrow {\left( {99} \right)^5} = {\left( {100 - 1} \right)^5} = {10^{10}} - 5 \times {10^8} + {10^7} - {10^5} + 500 - 1 = 9509900499 \\

\]

Hence, \[{\left( {99} \right)^5} = 9509900499\].

Note- These types of problems are solved by somehow converting the expression which needs to be evaluated into some form so that the binomial theorem or its special case are useful to obtain the answer.

Since, we have to find the value for ${\left( {99} \right)^5}$ which can be written as \[{\left( {100 - 1} \right)^5}\].

According to Binomial theorem, we know that

\[{\left( {x - 1} \right)^n} = {}^n{C_0}{x^n}{1^0} - {}^n{C_1}{x^{n - 1}}{1^1} + {}^n{C_2}{x^{n - 2}}{1^2} - ..... + {\left( { - 1} \right)^{n - 1}}{}^n{C_{n - 1}}{x^1}{1^{n - 1}} + {\left( { - 1} \right)^n}{}^n{C_n}{x^0}{1^n}\]

where \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] and \[{}^n{C_0} = 1,{}^n{C_1} = n,{}^n{C_{n - 1}} = n,{}^n{C_n} = 1\]

\[ \Rightarrow {\left( {x - 1} \right)^n} = {x^n} - n{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} - ..... + {\left( { - 1} \right)^{n - 1}}nx + {\left( { - 1} \right)^n}\]

In the above equation, put \[x = 100\] and \[n = 5\]

\[ \Rightarrow {\left( {99} \right)^5} = {\left( {100 - 1} \right)^5} = {100^5} - 5 \times {\left( {100} \right)^4} + {}^5{C_2}{\left( {100} \right)^3} - {}^5{C_3}{\left( {100} \right)^2} + 5 \times 100 - 1{\text{ }} \to {\text{(1)}}\]

Now, \[{}^5{C_2} = \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}} = \dfrac{{5!}}{{2!3!}} = \dfrac{{5 \times 4}}{2} = 10\] and \[{}^5{C_3} = \dfrac{{5!}}{{3!\left( {5 - 3} \right)!}} = \dfrac{{5!}}{{3!2!}} = \dfrac{{5 \times 4}}{2} = 10\]

Therefore, equation (1) becomes

\[

\Rightarrow {\left( {99} \right)^5} = {\left( {100 - 1} \right)^5} = {10^{10}} - 5 \times {10^8} + 10 \times {10^6} - 10 \times {10^4} + 500 - 1 \\

\Rightarrow {\left( {99} \right)^5} = {\left( {100 - 1} \right)^5} = {10^{10}} - 5 \times {10^8} + {10^7} - {10^5} + 500 - 1 = 9509900499 \\

\]

Hence, \[{\left( {99} \right)^5} = 9509900499\].

Note- These types of problems are solved by somehow converting the expression which needs to be evaluated into some form so that the binomial theorem or its special case are useful to obtain the answer.

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