
How do you use the Binomial theorem to find the value of \[{99^4}\] \[?\]
Answer
559.2k+ views
Hint: To find the value of the given \[{99^4}\] first we rewrite the 99 in addition or subtraction form then use the formula binomial expansion and the formula is given as \[{(a + b)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + ... + {}^n{C_n}{a^0}{b^n}\]. We apply the binomial expansion where n is 4. Further on simplification gives a required value.
Complete step-by-step answer:
Binomial theorem, states that for any positive integer n, the nth power of the sum of two numbers a and b may be expressed as the sum of n+1 terms of the form
\[(^n{C_r}){a^{n - r}}{b^r}\]
in the sequence of terms, the index r takes on the successive values 0, 1, 2, … ,n. The coefficients, called the binomial coefficients, are defined by the formula
\[ (^n{C_r}) = \dfrac{{n!}}{{(n - r)!r!}}\] in which n! (called n factorial) is the product of the first n natural numbers 1, 2, 3, … ,n (and where 0! is defined as equal to 1).
The formula of binomial expansion is
\[{(a + b)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + ... + {}^n{C_n}{a^0}{b^n}\]
Consider the given \[{99^4}\]
99 can be written as (100-1), then
\[ \Rightarrow {\left( {100 - 1} \right)^4}\]
Where a=100 and b=1, and n=4. using the formula of binomial expansion
\[ \Rightarrow {\left( {100 - 1} \right)^4} = {}^4{C_0}{\left( {100} \right)^4}{\left( { - 1} \right)^0} + {}^4{C_1}{\left( {100} \right)^3}{\left( { - 1} \right)^1} + {}^4{C_2}{\left( {100} \right)^2}{\left( { - 1} \right)^2} + {}^4{C_3}{\left( {100} \right)^1}{\left( { - 1} \right)^3} + {}^4{C_3}{\left( {100} \right)^0}{\left( { - 1} \right)^4}\]
Using the combination formula \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
Therefore, the value of \[^4{C_0} = 0\], \[^4{C_1} = 4\], \[^4{C_2} = 6\], \[^4{C_3} = 4\] and \[^4{C_4} = 1\], then
\[ \Rightarrow {\left( {100 - 1} \right)^4} = 1{\left( {100} \right)^4}.1 + 4{\left( {100} \right)^3}\left( { - 1} \right) + 6{\left( {100} \right)^2}1 + 4{\left( {100} \right)^1}\left( { - 1} \right) + 1.1.1\]
On simplification we have
\[ \Rightarrow {\left( {100 - 1} \right)^4} = {\left( {100} \right)^4} - 4{\left( {100} \right)^3} + 6{\left( {100} \right)^2} - 4{\left( {100} \right)^1} + 1\]
On further simplification we get
\[ \Rightarrow {\left( {100 - 1} \right)^4} = 100000000 - 4000000 + 60000 - 400 + 1\]
Adding all these terms we obtain
\[\therefore {\left( {99} \right)^4} = 96059601\]
Hence, the value of \[{99^4}\] using binomial theorem is 96059601.
So, the correct answer is “ 96059601”.
Note: To solve this type of this we use the binomial expansion formula and the formula is defined as \[{(a + b)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + ... + {}^n{C_n}{a^0}{b^n}\] by substituting the value of a, b and n we can calculate the solution for this question. On further simplification we obtain the required the solution
Complete step-by-step answer:
Binomial theorem, states that for any positive integer n, the nth power of the sum of two numbers a and b may be expressed as the sum of n+1 terms of the form
\[(^n{C_r}){a^{n - r}}{b^r}\]
in the sequence of terms, the index r takes on the successive values 0, 1, 2, … ,n. The coefficients, called the binomial coefficients, are defined by the formula
\[ (^n{C_r}) = \dfrac{{n!}}{{(n - r)!r!}}\] in which n! (called n factorial) is the product of the first n natural numbers 1, 2, 3, … ,n (and where 0! is defined as equal to 1).
The formula of binomial expansion is
\[{(a + b)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + ... + {}^n{C_n}{a^0}{b^n}\]
Consider the given \[{99^4}\]
99 can be written as (100-1), then
\[ \Rightarrow {\left( {100 - 1} \right)^4}\]
Where a=100 and b=1, and n=4. using the formula of binomial expansion
\[ \Rightarrow {\left( {100 - 1} \right)^4} = {}^4{C_0}{\left( {100} \right)^4}{\left( { - 1} \right)^0} + {}^4{C_1}{\left( {100} \right)^3}{\left( { - 1} \right)^1} + {}^4{C_2}{\left( {100} \right)^2}{\left( { - 1} \right)^2} + {}^4{C_3}{\left( {100} \right)^1}{\left( { - 1} \right)^3} + {}^4{C_3}{\left( {100} \right)^0}{\left( { - 1} \right)^4}\]
Using the combination formula \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
Therefore, the value of \[^4{C_0} = 0\], \[^4{C_1} = 4\], \[^4{C_2} = 6\], \[^4{C_3} = 4\] and \[^4{C_4} = 1\], then
\[ \Rightarrow {\left( {100 - 1} \right)^4} = 1{\left( {100} \right)^4}.1 + 4{\left( {100} \right)^3}\left( { - 1} \right) + 6{\left( {100} \right)^2}1 + 4{\left( {100} \right)^1}\left( { - 1} \right) + 1.1.1\]
On simplification we have
\[ \Rightarrow {\left( {100 - 1} \right)^4} = {\left( {100} \right)^4} - 4{\left( {100} \right)^3} + 6{\left( {100} \right)^2} - 4{\left( {100} \right)^1} + 1\]
On further simplification we get
\[ \Rightarrow {\left( {100 - 1} \right)^4} = 100000000 - 4000000 + 60000 - 400 + 1\]
Adding all these terms we obtain
\[\therefore {\left( {99} \right)^4} = 96059601\]
Hence, the value of \[{99^4}\] using binomial theorem is 96059601.
So, the correct answer is “ 96059601”.
Note: To solve this type of this we use the binomial expansion formula and the formula is defined as \[{(a + b)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + ... + {}^n{C_n}{a^0}{b^n}\] by substituting the value of a, b and n we can calculate the solution for this question. On further simplification we obtain the required the solution
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