Question

How do you expand ${{\left( x-3 \right)}^{5}}$ using Pascal’s Triangle?

Answer 1

These types of binomial problems are very easy to solve once we can figure out how to expand a series. We can expand a particular given series in many ways. We can do it using the binomial expansion and we can also do it using the method of Pascal’s Triangle. To solve it using the Pascal’s Triangle method, we first of all need to have a complete idea about how and where we can use this method for expansions. Pascal’s Triangle is a triangular array that is constructed by adding the adjacent terms in the previous row. In this problem since the expansion is given as the power of $5$ , we need to see the sixth row of the Pascal’s Triangle and then apply it to our problem.

Complete step-by-step solution:
Now we start off with the solution to the problem by writing of the Pascal’s Triangle for the first 6 terms. They are written as,

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{1st Row} \\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{2nd Row} \\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{3rd Row} \\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{4th Row} \\\ &\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\text{5th Row} \\\ & 1\,\,\,\,\,\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,\,\,\,\,\,\,10\,\,\,\,\,\,\,\,\,\,\,\,\,10\,\,\,\,\,\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\text{6th Row} \\\ \end{aligned}$$ So, in this problem, what we need is the sixth row, to find all the terms of the expansion of $${{\left( x-3 \right)}^{5}}$$ . Now we write all the terms according to the sixth row as, $$\Rightarrow {{\left( x-3 \right)}^{5}}={{x}^{5}}+5{{x}^{4}}\left( -3 \right)+10{{x}^{3}}{{\left( -3 \right)}^{2}}+10{{x}^{2}}{{\left( -3 \right)}^{3}}+5x{{\left( -3 \right)}^{4}}+{{\left( -3 \right)}^{5}}$$ Now what is left is evaluating the terms of the formed expansion, $$\Rightarrow {{\left( x-3 \right)}^{5}}={{x}^{5}}-15{{x}^{4}}+90{{x}^{3}}-270{{x}^{2}}+405x-243$$ This gives us our required answer. **Note:** Solving these types of problems is not a big deal once we know about the law of the Pascal’s Triangle and how to implement it. One very important thing to note down here is that if we want to find the expansion of the ${{n}^{th}}$ term then we must consider the ${{\left( n+1 \right)}^{th}}$ row of the Pascal’s Triangle. However, we can also solve this equation using the normal binomial expansion which we are accustomed to. Pascal’s Triangle implementation is however a much simpler implementation.