Question

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

Answer 1

Hint: If the expression is in the ${{\left( Ax+B \right)}^{n}}$, this it is expanded by the Binomial theorem,
Therefore, the ${{x}^{k}}$ term is
$\left( n,k \right).{{\left( Ax \right)}^{k}}.{{B}^{n-k}}$
$=\left( n,k \right).{{A}^{k}}.{{B}^{n-k}}.{{x}^{k}}$
Or in other case the equivantly; the coefficient of ${{x}^{k}}$ will be $\left( n,k \right).{{A}^{k}}.{{B}^{n-k}}$
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 ${{\left( 2a+1 \right)}^{5}}$
Here, The expression is in the ${{\left( Ax+B \right)}^{n}}$ form.
So, we have to expand it by using the Binomial theorem.
${{\left( x+y \right)}^{n}}=\sum\limits_{k=}^{n}{\left( \dfrac{n}{k} \right){{x}^{n-k}}{{y}^{k}}}$
We want the ${{a}^{2}}$ term,
So, if $x=a$ then we need $n-k=2$
Since,
$n=5$ implies $k=3$
Then, $\left( \dfrac{5}{3} \right){{\left( 2a \right)}^{5-3}}{{\left( 1 \right)}^{3}}$
$\dfrac{5!}{3!2!}.{{\left( 2a \right)}^{2}}.{{\left( 1 \right)}^{3}}$
$\dfrac{5\times 4\times \left( 3\times 2\times 1 \right)}{\left( 3\times 2\times 1 \right)\left( 2\times 1 \right)}\times {{\left( 4a \right)}^{2}}$
$\dfrac{20}{2}\times {{\left( 4a \right)}^{2}}=40{{a}^{2}}$
Hence,
The coefficient for the expansion $\left( 2a+1 \right)$ is $40{{a}^{2}}.$ 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.

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 ${{\left( x+y \right)}^{n}}$ Also while expanding check its degree of each term. We will also use other methods such as Pascal’s triangle for determining the solution. So this is important tips for solving the binomial theorem problems.