Problem 37 - Algebra and Trigonometry

Question

Write the first three terms in each binomial expansion, expressing the result in simplified form. $$ \left(y^{3}-1\right)^{20} $$

Step-by-Step Solution

Answer

The first three terms in the binomial expansion of \((y^{3}-1)^{20}\) are \(y^{60}\), \(-20y^{57}\), and \(190y^{54}\).

Step 1 - Identify a and b

In this given example, the binomial expression is \((y^{3}-1)^{20}\). Let's identify 'a' and 'b' in the binomial theorem. Here, \(a = y^3\) and \(b = -1\).

Step 2 - Apply binomial theorem for the first term

We plug \(k=0\) into the formula \({20 \choose k} a^{20-k} b^{k}\), where \( n = 20\), as \(k=0\), \(a=y^3\) and \(b=-1\)\. This gives: \({20 \choose 0} (y^3)^{20-0} (-1)^{0}\), which simplifies to \(1 * y^{60} * 1 = y^{60}\). So, the first term of the expansion is \(y^{60}\).

Step 3 - Apply binomial theorem for the Second term

We now plug \(k=1\) into the formula \({20 \choose k} a^{20-k} b^{k}\), again for \( n = 20\), \(k=1\), \(a=y^3\) and \(b=-1\)\. This gives: \({20 \choose 1} (y^3)^{20-1} (-1)^{1}\), which simplifies to \(-20 * y^{57}\). So, the second term of the expansion is \(-20y^{57}\).

Step 4 - Apply binomial theorem for the Third term

This time plug \(k=2\) into the formula \({20 \choose k} a^{20-k} b^{k}\), again for \( n = 20\), \(k=2\), \(a=y^3\) and \(b=-1\)\. This results: \({20 \choose 2} (y^3)^{20-2} (-1)^{2}\), which simplifies to \(190 * y^{54}\). So, the third term of the expansion is \(190y^{54}\).