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Question: How do you find the coefficient of \({{a}^{2}}\) in the expansion of \({{\left( 2x+1 \right)}^{5}}?\...

How do you find the coefficient of a2{{a}^{2}} in the expansion of (2x+1)5?{{\left( 2x+1 \right)}^{5}}?

Explanation

Solution

If the expression is in the (Ax+B)n{{\left( Ax+B \right)}^{n}} This is expanded by the Binomial theorem,
Therefore the xk{{x}^{k}} term is
C(n,k).(Ax)k.BnkC\left( n,k \right).{{\left( Ax \right)}^{k}}.{{B}^{n-k}}
=C(n,k).Ak.Bnk.xk=C\left( n,k \right).{{A}^{k}}.{{B}^{n-k}}.{{x}^{k}}
Or in other case the equivalently the coefficient of xk{{x}^{k}} will be C(n.k).Ak.BnkC\left( n.k \right).{{A}^{k}}.{{B}^{n-k}}
Then, take the given coefficient and solve it by the above given condition for getting the value.

Complete step by step solution:
The given expression is (2a+1)5{{\left( 2a+1 \right)}^{5}}
Here the expression is in the (Ax+B)n{{\left( Ax+B \right)}^{n}} form.
So, we have to expand it by using binomial theorem.
(x+y)n=k=0n(nk)xnkyk{{\left( x+y \right)}^{n}}=\sum\limits_{k=0}^{n}{\left( \dfrac{n}{k} \right){{x}^{n-k}}{{y}^{k}}}
We want the a2{{a}^{2}} term.
So, if x=20x=20 then we need nk=2.n-k=2.
Since n=5n=5 implies k=3k=3
Then, (53)(2a)53(1)3\left( \dfrac{5}{3} \right){{\left( 2a \right)}^{5-3}}{{\left( 1 \right)}^{3}}
5!3!2!×(2C)2×(1)3\dfrac{5!}{3!2!}\times {{\left( 2C \right)}^{2}}\times {{\left( 1 \right)}^{3}}
5×4×(3×2×1)×(4a)2(3×2×1)(2×1)\Rightarrow \dfrac{5\times 4\times \left( 3\times 2\times 1 \right)\times {{\left( 4a \right)}^{2}}}{\left( 3\times 2\times 1 \right)\left( 2\times 1 \right)}
202×(4a)2=40a2\Rightarrow \dfrac{20}{2}\times {{\left( 4a \right)}^{2}}=40{{a}^{2}}
Hence,
The coefficient for the expansion (2a+1)\left( 2a+1 \right) is 40a240{{a}^{2}}
We will also find out by using Pascal’s triangle or you can also use the other method but this is the easiest way for determining the value.

Additional Information:
The Binomial theorem is that for any positive n'n' integer the power of sum of two numbers that is aa and bb it is expressed as n+1n+1 term. In the sequence term the index rr takes on the successive value 0,1,2,3...n0,1,2,3...n then the coefficient is known as binomial coefficient which is defined by formula.
(nr)=n!(np)!p!\left( \dfrac{n}{r} \right)=n!|\left( n-p \right)!p! where n!n! (called as the factorial of nn) is also the product of first n.n.
Natural number 1,2,3...n.1,2,3...n. The other method is also for determining the coefficient is the Pascal’s triangle. If is a triangle arrangement of the number which gives us expansion of coefficient of coefficient of any expression such as (x+g)?\left( x+g \right)?

Note: We should check that the term is in which form. We have the expression in binomial because it has two terms. So, we have to expand with (x+y)n{{\left( x+y \right)}^{n}}Also while the expander checks its degree of each term. We will also use other methods such as Pascal’s triangle for determining the solution. So, these are important tips for solving the binomial theorem problems.