
How do I use pascals’ triangle to expand the binomial \[{{\left( d-3 \right)}^{6}}\] ?
Answer
543k+ views
Hint: To solve the above question we have to apply pascal's triangle. Through the Pascal triangle, we will get the coefficient of the binomial expansion. As every row of pascal’s triangle is built from the row above it. The binomial theorem tells us that we can use these coefficients to find the entire expanded series.
Complete step by step answer:
This question belongs to the concept of the binomial theorem and Pascal's triangle expansion. It is one of the most interesting number patterns. Pascal’s triangle is basically an array of the binomial coefficients that come in algebra and probability theory. By using its coefficient, we can find the expanded binomial theorem. The pascal’s triangle always starts with 1. Since, we can only find the coefficients of the term therefore, in order to find the signs, we will use binomial expansion.
For the above question first, we will write the pascal’s triangle after that we will use it to get the coefficients. The given binomial is \[{{\left( d-3 \right)}^{6}}\] .
The pascal’s triangle is
\[\begin{array}{*{35}{l}}
1 \\
1\text{ }1 \\
1\text{ }2\text{ }1 \\
1\text{ }3\text{ }3\text{ }1 \\
1\text{ }4\text{ }6\text{ }4\text{ }1 \\
1\text{ }5\text{ }10\text{ }10\text{ }5\text{ }1 \\
1\text{ }6\text{ }15\text{ }20\text{ }15\text{ }6\text{ }1 \\
\end{array}\]
We will use the numbers from the triangle as coefficients for the expansion.
\[{{(d-3)}^{6}}=1{{d}^{6}}{{(-3)}^{0}}+6{{d}^{5}}{{(-3)}^{1}}+15{{d}^{4}}{{(-3)}^{2}}+20{{d}^{3}}{{(-3)}^{3}}+15{{d}^{2}}{{(-3)}^{4}}+6{{d}^{1}}{{(-3)}^{5}}+1{{d}^{0}}{{(-3)}^{6}}\]
\[{{(d-3)}^{6}}={{x}^{6}}-18{{x}^{5}}+135{{x}^{4}}-540{{x}^{3}}+1215{{x}^{2}}-1458x+729\]
Therefore, the expansion for \[{{\left( d-3 \right)}^{6}}\] is \[{{x}^{6}}-18{{x}^{5}}+135{{x}^{4}}-540{{x}^{3}}+1215{{x}^{2}}-1458x+729\] .
Note:
The pascal’s triangle is never-ending which means infinite and there is no bottom side. Carefully write the coefficients from the pascals triangle for the binomial expansion. Also, we can’t determine the signs of the terms through the pascal triangle. Keep in mind the steps followed for future use.
Complete step by step answer:
This question belongs to the concept of the binomial theorem and Pascal's triangle expansion. It is one of the most interesting number patterns. Pascal’s triangle is basically an array of the binomial coefficients that come in algebra and probability theory. By using its coefficient, we can find the expanded binomial theorem. The pascal’s triangle always starts with 1. Since, we can only find the coefficients of the term therefore, in order to find the signs, we will use binomial expansion.
For the above question first, we will write the pascal’s triangle after that we will use it to get the coefficients. The given binomial is \[{{\left( d-3 \right)}^{6}}\] .
The pascal’s triangle is
\[\begin{array}{*{35}{l}}
1 \\
1\text{ }1 \\
1\text{ }2\text{ }1 \\
1\text{ }3\text{ }3\text{ }1 \\
1\text{ }4\text{ }6\text{ }4\text{ }1 \\
1\text{ }5\text{ }10\text{ }10\text{ }5\text{ }1 \\
1\text{ }6\text{ }15\text{ }20\text{ }15\text{ }6\text{ }1 \\
\end{array}\]
We will use the numbers from the triangle as coefficients for the expansion.
\[{{(d-3)}^{6}}=1{{d}^{6}}{{(-3)}^{0}}+6{{d}^{5}}{{(-3)}^{1}}+15{{d}^{4}}{{(-3)}^{2}}+20{{d}^{3}}{{(-3)}^{3}}+15{{d}^{2}}{{(-3)}^{4}}+6{{d}^{1}}{{(-3)}^{5}}+1{{d}^{0}}{{(-3)}^{6}}\]
\[{{(d-3)}^{6}}={{x}^{6}}-18{{x}^{5}}+135{{x}^{4}}-540{{x}^{3}}+1215{{x}^{2}}-1458x+729\]
Therefore, the expansion for \[{{\left( d-3 \right)}^{6}}\] is \[{{x}^{6}}-18{{x}^{5}}+135{{x}^{4}}-540{{x}^{3}}+1215{{x}^{2}}-1458x+729\] .
Note:
The pascal’s triangle is never-ending which means infinite and there is no bottom side. Carefully write the coefficients from the pascals triangle for the binomial expansion. Also, we can’t determine the signs of the terms through the pascal triangle. Keep in mind the steps followed for future use.
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