
Find the coefficient of ${{a}^{2}}{{b}^{2}}$ in the expansion of ${{\left( a+b \right)}^{18}}$.
Answer
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Hint: Use binomial theorem to find the expansion of the given binomial term. Find the general term of ${{\left( a+b \right)}^{18}}$ by using the formula: ${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$. Substitute the value of $n\text{ and }r$ to find the coefficient of ${{a}^{2}}{{b}^{2}}$.
Complete step-by-step answer:
In elementary algebra, the binomial theorem or binomial expansion describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial ${{(x+y)}^{k}}$ into a sum involving terms of the form $p{{x}^{m}}{{y}^{n}}$, where the exponents $m\text{ and }n$ are non-negative integers with $m+n=k$, and the coefficient $p$ of each term is a specific positive integer depending on $k\text{ and }m$. The coefficient $p$ in the form of $p{{x}^{m}}{{y}^{n}}$ is known as the binomial coefficient ${}^{k}{{C}_{m}}\,\text{ or }{}^{k}{{C}_{n}}$. These coefficients for varying $k\text{ and }m$can be arranged to form Pascal’s triangle. For example:
${{(x+y)}^{4}}={{x}^{4}}+4{{x}^{3}}y+6{{x}^{2}}{{y}^{2}}+4x{{y}^{3}}+{{y}^{4}}$.
Now, let us come to the question. We have to find the coefficient of ${{a}^{2}}{{b}^{2}}$ in the expansion of ${{\left( a+b \right)}^{18}}$. Since, general term of an binomial expression is given as, ${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$, here $n$ is the total power of the given expression. Using this for ${{\left( a+b \right)}^{18}}$ we get,
${{\left( a+b \right)}^{18}}={}^{18}{{C}_{r}}{{a}^{18-r}}{{b}^{r}}$
Now, if we will substitute $r=2$, then we will get the ${{b}^{2}}$ term but with this we will get ${{a}^{16}}$ and if we will substitute $a=16$ then we will get ${{a}^{2}}$ term but with this we will get ${{b}^{16}}$. Hence, we can conclude that there is no term like ‘${{a}^{2}}{{b}^{2}}$’.
Therefore, the coefficient of ${{a}^{2}}{{b}^{2}}$ is $0$.
Note: We have an easy method by which we can say that the coefficient of ${{a}^{2}}{{b}^{2}}$ is $0$. Note that “the sum of powers of the two terms in the expanded form of the binomial is the total power of the expression”. In our question, the total power of the expression ${{\left( a+b \right)}^{18}}$ is 18 and the sum of the powers in ${{a}^{2}}{{b}^{2}}$ is 4. Hence, no such term exists. Therefore, the coefficient of this term must be $0$.
Complete step-by-step answer:
In elementary algebra, the binomial theorem or binomial expansion describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial ${{(x+y)}^{k}}$ into a sum involving terms of the form $p{{x}^{m}}{{y}^{n}}$, where the exponents $m\text{ and }n$ are non-negative integers with $m+n=k$, and the coefficient $p$ of each term is a specific positive integer depending on $k\text{ and }m$. The coefficient $p$ in the form of $p{{x}^{m}}{{y}^{n}}$ is known as the binomial coefficient ${}^{k}{{C}_{m}}\,\text{ or }{}^{k}{{C}_{n}}$. These coefficients for varying $k\text{ and }m$can be arranged to form Pascal’s triangle. For example:
${{(x+y)}^{4}}={{x}^{4}}+4{{x}^{3}}y+6{{x}^{2}}{{y}^{2}}+4x{{y}^{3}}+{{y}^{4}}$.
Now, let us come to the question. We have to find the coefficient of ${{a}^{2}}{{b}^{2}}$ in the expansion of ${{\left( a+b \right)}^{18}}$. Since, general term of an binomial expression is given as, ${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$, here $n$ is the total power of the given expression. Using this for ${{\left( a+b \right)}^{18}}$ we get,
${{\left( a+b \right)}^{18}}={}^{18}{{C}_{r}}{{a}^{18-r}}{{b}^{r}}$
Now, if we will substitute $r=2$, then we will get the ${{b}^{2}}$ term but with this we will get ${{a}^{16}}$ and if we will substitute $a=16$ then we will get ${{a}^{2}}$ term but with this we will get ${{b}^{16}}$. Hence, we can conclude that there is no term like ‘${{a}^{2}}{{b}^{2}}$’.
Therefore, the coefficient of ${{a}^{2}}{{b}^{2}}$ is $0$.
Note: We have an easy method by which we can say that the coefficient of ${{a}^{2}}{{b}^{2}}$ is $0$. Note that “the sum of powers of the two terms in the expanded form of the binomial is the total power of the expression”. In our question, the total power of the expression ${{\left( a+b \right)}^{18}}$ is 18 and the sum of the powers in ${{a}^{2}}{{b}^{2}}$ is 4. Hence, no such term exists. Therefore, the coefficient of this term must be $0$.
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