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

Last updated date: 21st Jul 2024
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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.

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.