How do you use the binomial ${(x + 2y)^5}$ using Pascal’s triangle?

Hint: As we know that we have to apply the pascal’s triangle to expand the binomial. We know that it is an infinite equilateral triangle which consists of a sequence of numbers. It starts with $1$. The second row consists of the sum of two numbers above it, Similarly we can find out the values of the next rows.

Complete step-by-step answer:
As we know that the main application of this triangle is to solve a binomial function. If the binomial equation is ${(a + b)^n}$, then the expansion is ${C_1}{a^n}{b^0} + {C_2}{a^{n - 1}}{b^1} + ... + {C_n}{a^0}{b^n}$.
The above equation represents the binomial expansion formula.
But here we are not going to solve this question by the binomial expansion.
Pascal's triangle is a triangular array constructed by summing adjacent elements in preceding rows. Pascal's triangle contains the values of the binomial coefficient.
The pascal’s triangle is given by
\[\begin{array}{*{20}{c}}
  {{{(x + y)}^0} = } \\
  {{{(x + y)}^1} = } \\
  {{{(x + y)}^2} = } \\
  {{{(x + y)}^3} = } \\
  {{{(x + y)}^4} = } \\
  {{{(x + y)}^5} = }
\end{array}\begin{array}{*{20}{c}}
  {}&{}&{}&{}&{}&1&{}&{}&{}&{}&{} \\
  {}&{}&{}&{}&1&{}&1&{}&{}&{}&{} \\
  {}&{}&{}&1&{}&2&{}&1&{}&{}&{} \\
  {}&{}&1&{}&3&{}&3&{}&1&{}&{} \\
  {}&1&{}&4&{}&6&{}&4&{}&1&{} \\
  1&{}&5&{}&{10}&{}&{10}&{}&5&{}&1
\end{array}\]
The above pascal’s triangle is given till 5th power.
Since we have $n = 5$, and from the above triangle we can see that the fifth order term is $1,5,10,10,5,1$.
For this we have the formula, ${(a + b)^5} = 1{a^5}{b^0} + 5{a^4}{b^1} + 10{a^3}{b^2} + 10{a^2}{b^3} + 5{x^1}{b^4} + 1{a^0}{b^5}$.
By applying the above formula we can write ${(x + 2y)^5} = 1{(x)^5}{(2y)^0} + 5{(x)^4}{(2y)^1} + 10{(x)^3}{(2y)^2} + 10{(x)^2}{(2y)^3} + 5{(x)^1}{(2y)^4} + 1{(x)^0}{(2y)^5}$.
On multiplying
It gives us the expression: ${x^5} + 10{x^4}{y^{}} + 40{x^3}{y^2} + 80{x^2}{y^3} + 80x{y^4} + 32{y^5}$.
Hence the required value is ${x^5} + 10{x^4}{y^{}} + 40{x^3}{y^2} + 80{x^2}{y^3} + 80x{y^4} + 32{y^5}$.
So, the correct answer is “${x^5} + 10{x^4}{y^{}} + 40{x^3}{y^2} + 80{x^2}{y^3} + 80x{y^4} + 32{y^5}$”.

Note: We should note that pascal’s triangle is helpful only when the value of $n$ is small in the equation ${(a + b)^n}$. If the value is large then it is very tedious to draw the triangle until we reach $n$. The formula that we used above because the question has the positive sign i.e. sum of the values.