
How to use Pascal’s triangle to expand a binomial?
Answer
542.4k+ views
Hint: The formula for Pascal’s triangle comes from a relationship of coefficients. The demonstration illustrates the pattern. Pascal’s triangle presents a formula that allows creating the coefficients of the term in a binomial expansion.
Complete step by step answer:
Pascal’s triangle provides a formula for expanding binomials.
\[
\Rightarrow {\left( {x + y} \right)^0} = 1 \\
\Rightarrow {\left( {x + y} \right)^1} = 1x + 1y \\
\Rightarrow {\left( {x + y} \right)^2} = 1{x^2} + 2xy + 1{y^2} \\
\Rightarrow {\left( {x + y} \right)^3} = 1{x^3} + 3{x^2}y + 3x{y^2} + 1{y^3} \\
\Rightarrow {\left( {x + y} \right)^4} = 1{x^4} + 4{x^3}y + 6{x^2}{y^2} + 4x{y^3} + 1{y^4} \\
\]
We need to be able to recognize the patterns that are generated when we expand the binomial\[\left( {x + y} \right)\]. The first two rows shows that anything to the power 0 is equal to 1 and anything raised to the first power is just equal to itself.
However, we are including the coefficients here because when we expand the quantity \[{\left( {x + y} \right)^2}\] it can be noticed that the first terms has 2 factors of x, which matches the power and then the next term the power on the x decreases to 1. Then the factors of y are picked and when the exponents are summed, it is always going to equal the original exponent on the binomial. In the third term we have 0 factors of x but we have 2 factors of y that are y squared. Again, it can be noticed that each of the exponents always add to 2, which is the exponent of the binomial. Similarly in the third row we have the quantity \[{\left( {x + y} \right)^3}\] and x is started with 3 factors and the same process continues.
Note: The coefficients actually make up what is called Pascal’s triangle and there is a special way to generate these numbers by starting with a triangle of ones. Each row starts with a 1 and ends with a 1 and the terms in the middle are formed by adding the two terms above it. The patterns are combined with the variables with Pascal's triangle to expand binomials. The exponents in the original binomial tell us which row we go to in Pascal’s triangle to find the coefficients.
Complete step by step answer:
Pascal’s triangle provides a formula for expanding binomials.
\[
\Rightarrow {\left( {x + y} \right)^0} = 1 \\
\Rightarrow {\left( {x + y} \right)^1} = 1x + 1y \\
\Rightarrow {\left( {x + y} \right)^2} = 1{x^2} + 2xy + 1{y^2} \\
\Rightarrow {\left( {x + y} \right)^3} = 1{x^3} + 3{x^2}y + 3x{y^2} + 1{y^3} \\
\Rightarrow {\left( {x + y} \right)^4} = 1{x^4} + 4{x^3}y + 6{x^2}{y^2} + 4x{y^3} + 1{y^4} \\
\]
We need to be able to recognize the patterns that are generated when we expand the binomial\[\left( {x + y} \right)\]. The first two rows shows that anything to the power 0 is equal to 1 and anything raised to the first power is just equal to itself.
However, we are including the coefficients here because when we expand the quantity \[{\left( {x + y} \right)^2}\] it can be noticed that the first terms has 2 factors of x, which matches the power and then the next term the power on the x decreases to 1. Then the factors of y are picked and when the exponents are summed, it is always going to equal the original exponent on the binomial. In the third term we have 0 factors of x but we have 2 factors of y that are y squared. Again, it can be noticed that each of the exponents always add to 2, which is the exponent of the binomial. Similarly in the third row we have the quantity \[{\left( {x + y} \right)^3}\] and x is started with 3 factors and the same process continues.
Note: The coefficients actually make up what is called Pascal’s triangle and there is a special way to generate these numbers by starting with a triangle of ones. Each row starts with a 1 and ends with a 1 and the terms in the middle are formed by adding the two terms above it. The patterns are combined with the variables with Pascal's triangle to expand binomials. The exponents in the original binomial tell us which row we go to in Pascal’s triangle to find the coefficients.
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