Question

How do you expand \[{{\left( x+y \right)}^{6}}\]using pascal’s triangle?

Answer 1

Hint: In the given question, we have been asked to expand \[{{\left( x+y \right)}^{6}}\] by using Pascal’s triangle. The row of pascal’s triangle provides the coefficient to \[{{\left( a+b \right)}^{n}}\]. The triangle can be used to calculate the coefficients of the expansion of \[{{\left( a+b \right)}^{n}}\] by taking the exponent \[n\] and adding 1.

Complete step by step answer:
The binomial expansion follows the rule, i.e.
\[{{\left( a+b \right)}^{n}}={{c}_{0}}{{a}^{n}}{{b}^{0}}+{{c}_{1}}{{a}^{n-1}}{{b}^{1}}+......+{{c}_{n-1}}{{a}^{1}}{{b}^{n-1}}+{{c}_{n}}{{a}^{0}}{{b}^{n}}\] 
Consider the \[{{n}^{th}}\] power of the binomial expression\[\left( a+b \right)\], namely\[{{\left( a+b \right)}^{n}}\]. It must be a polynomial in \[a\] and \[b\] of degree \[n\] , which means that the exponent of \[a\] and \[b\] must sum to\[n\].
The given binomial expression is,
\[{{\left( x+y \right)}^{6}}={{c}_{0}}{{x}^{6}}{{y}^{0}}+{{c}_{1}}{{x}^{6-1}}{{y}^{1}}+{{c}_{2}}{{x}^{6-2}}{{y}^{2}}+{{c}_{3}}{{x}^{6-3}}{{y}^{3}}+{{c}_{4}}{{x}^{6-4}}{{y}^{4}}+{{c}_{5}}{{x}^{6-5}}{{y}^{5}}+{{c}_{6}}{{x}^{6-6}}{{y}^{6}}\]
Pascal’s triangle will be represented as:
1
1−1
1−2−1
1−3−3−1
1−4−6−4−1
1−5−10−10−5−1
1−6−15−20−15−6−1

For \[{{\left( x+y \right)}^{6}},\ where\ n=6\], so the coefficients of the expansion will correspond with row 7th.
The 7^th row of the pascal’s triangle will be the value of the coefficients
According to the given binomial expression, the expansion will be as follows:
\[{{\left( x+y \right)}^{6}}={{c}_{0}}{{x}^{6}}{{y}^{0}}+{{c}_{1}}{{x}^{6-1}}{{y}^{1}}+{{c}_{2}}{{x}^{6-2}}{{y}^{2}}+{{c}_{3}}{{x}^{6-3}}{{y}^{3}}+{{c}_{4}}{{x}^{6-4}}{{y}^{4}}+{{c}_{5}}{{x}^{6-5}}{{y}^{5}}+{{c}_{6}}{{x}^{6-6}}{{y}^{6}}\]
 The values of the coefficients, from the pascal’s triangle, are 1−6−15−20−15−6−1
Substitute the value of coefficient from the pascal’s triangle in the above expansion in a order, we get
\[{{\left( x+y \right)}^{6}}=1\left( {{x}^{6}} \right)\left( {{y}^{0}} \right)+6\left( {{x}^{6-1}} \right)\left( {{y}^{1}} \right)+15\left( {{x}^{6-2}} \right)\left( {{y}^{2}} \right)+20\left( {{x}^{6-3}} \right)\left( {{y}^{3}} \right)+15\left( {{x}^{6-4}} \right)\left( {{y}^{4}} \right)+6\left( {{x}^{6-5}} \right)\left( {{y}^{5}} \right)+1\left( {{x}^{6-6}} \right)\left( {{y}^{6}} \right)\]
Simplify the numbers that is in the power form, we get
\[{{\left( x+y \right)}^{6}}=1\left( {{x}^{6}} \right)\left( {{y}^{0}} \right)+6\left( {{x}^{5}} \right)\left( {{y}^{1}} \right)+15\left( {{x}^{4}} \right)\left( {{y}^{2}} \right)+20\left( {{x}^{3}} \right)\left( {{y}^{3}} \right)+15\left( {{x}^{2}} \right)\left( {{y}^{4}} \right)+6\left( {{x}^{1}} \right)\left( {{y}^{5}} \right)+1\left( {{x}^{0}} \right)\left( {{y}^{6}} \right)\]Anything raised to power \[0\]is equals to\[1\], we get
\[{{\left( x+y \right)}^{6}}=1\left( {{x}^{6}} \right)\left( 1 \right)+6\left( {{x}^{5}} \right)\left( {{y}^{1}} \right)+15\left( {{x}^{4}} \right)\left( {{y}^{2}} \right)+20\left( {{x}^{3}} \right)\left( {{y}^{3}} \right)+15\left( {{x}^{2}} \right)\left( {{y}^{4}} \right)+6\left( {{x}^{1}} \right)\left( {{y}^{5}} \right)+1\left( 1 \right)\left( {{y}^{6}} \right)\]
Simplify the expression further, we get
\[{{\left( x+y \right)}^{6}}={{x}^{6}}+6{{x}^{5}}y+15{{x}^{4}}{{y}^{2}}+20{{x}^{3}}{{y}^{3}}+15{{x}^{2}}{{y}^{4}}+6x{{y}^{5}}+{{y}^{6}}\]
Therefore, this is the expansion of the binomial expression using Pascal's triangle.

Note: Each number in the pascal’s triangle is the sum of the two above, each of which counts the number of ways to get that point by a sequence of left and right choices. So all the numbers in Pascal's triangle count the number of ways to reach them by left or right choices starting at the top. The row of pascal’s triangle provides the coefficient to \[{{\left( a+b \right)}^{n}}\]. The triangle can be used to calculate the coefficients of the expansion of \[{{\left( a+b \right)}^{n}}\] by taking the exponent \[n\] and adding 1.