Question

the binomial theorem
the first term of this binomial is 2x.
the second term of this binomial is (-y^2).
what is the correct formula for the fifth term in this binomial expansion?
choose the simplified form of the fifth term of the expansion.
\\(^6c_5(2x)(-y^2)^5\\)
\\(^6c_4(2x)^4(-y^2)^2\\)
\\(^6c_3(2x)^3(-y^2)^3\\)
\\(-30x^2y^8\\)
\\(30x^2y^8\\)
\\(60x^2y^8\\)

Explanation:

Step1: Recall Binomial Theorem

The binomial theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where the \((r + 1)\)-th term is given by \(T_{r+1}=\binom{n}{r}a^{n - r}b^{r}\).

Step2: Identify \(n\), \(a\), \(b\) and \(r\) for the fifth term

We know the first term is \(2x\) (when \(r = 0\), \(T_{1}=\binom{n}{0}a^{n}b^{0}=a^{n}=2x\)) and the second term is \(-y^{2}\) (when \(r = 1\), \(T_{2}=\binom{n}{1}a^{n - 1}b^{1}\)). From the options for the formula of the fifth term, we can see \(n = 6\) (since we have \(\binom{6}{r}\)). For the fifth term, \(r+1 = 5\), so \(r = 4\). Here, \(a = 2x\) and \(b=-y^{2}\).

Step3: Apply the binomial term formula

Using \(T_{r + 1}=\binom{n}{r}a^{n - r}b^{r}\) with \(n = 6\), \(r = 4\), \(a = 2x\), \(b=-y^{2}\), we get \(T_{5}=\binom{6}{4}(2x)^{6 - 4}(-y^{2})^{4}\)? Wait, no, wait. Wait, let's re - check. Wait, the first term: if the first term is \(2x\), and the second term is \(-y^{2}\), maybe the binomial is \((2x - y^{2})^{6}\)? Let's check the term numbering. The general term is \(T_{k+1}=\binom{6}{k}(2x)^{6 - k}(-y^{2})^{k}\). For the fifth term, \(k + 1=5\), so \(k = 4\). So \(T_{5}=\binom{6}{4}(2x)^{6 - 4}(-y^{2})^{4}\)? Wait, no, the options have \(\binom{6}{4}(2x)^{4}(-y^{2})^{2}\)? Wait, maybe I made a mistake. Wait, let's look at the options for the formula. The options are \(\binom{6}{5}(2x)(-y^{2})^{5}\), \(\binom{6}{4}(2x)^{4}(-y^{2})^{2}\), \(\binom{6}{2}(2x)^{2}(-y^{2})^{4}\). Wait, for the fifth term, the index \(r\) (where term is \(T_{r}\)): if the first term is \(T_{1}\), then \(T_{5}\) corresponds to \(r = 4\) in \(T_{r+1}\) or \(r = 4\) in \(T_{r}\) (depending on numbering). Wait, let's calculate the exponent. Let's assume the binomial is \((2x - y^{2})^{6}\). The general term is \(T_{k + 1}=\binom{6}{k}(2x)^{6 - k}(-y^{2})^{k}\). For \(k=4\) (since \(k + 1 = 5\) for the fifth term), we have \(T_{5}=\binom{6}{4}(2x)^{6 - 4}(-y^{2})^{4}\)? No, wait \(6-4 = 2\), so \((2x)^{2}\), and \((-y^{2})^{4}\). But wait, the options: one of the options is \(\binom{6}{4}(2x)^{4}(-y^{2})^{2}\)? Wait, maybe I messed up \(a\) and \(b\). Wait, the first term is \(2x\), which is when \(k = 0\): \(T_{1}=\binom{6}{0}(2x)^{6}(-y^{2})^{0}= (2x)^{6}\), but the problem says the first term is \(2x\). Oh! So maybe the binomial is \((2x - y^{2})^{6}\) is wrong. Wait, the first term is \(2x\), so the binomial must be of the form \((2x + c)^{n}\), and the second term is \(-y^{2}\). Wait, maybe the binomial is \((2x - y^{2})^{6}\), but the first term is \(\binom{6}{0}(2x)^{6}=64x^{6}\), but the problem says the first term is \(2x\). So there must be a mistake in my initial assumption. Wait, the problem says "The first term of this binomial is \(2x\). The second term of this binomial is \(-y^{2}\)". So maybe the binomial is \((2x - y^{2})^{n}\), and when \(n = 6\)? No, the first term of \((a + b)^n\) is \(a^{n}\). So if the first term is \(2x\), then \(a^{n}=2x\), which implies \(n = 1\), but that can't be. Wait, maybe the binomial is \((2x - y^{2})^{6}\) and the first term is mis - described? Or maybe the "first term" here is the first term in the expansion of \((2x - y^{2})^{6}\) when we consider the terms with respect to the powers? No, that doesn't make sense. Wait, looking at the options for the formula of the fifth term, we have \(\binom{6}{4}(2x)^{4}(-y^{2})^{2}\). Let's calculate \(\binom{6}{4}=\frac{6!}{4!(6 - 4)!}=\frac{6\times5}{2\times1}=15\). \((2x)^{4}=16x^{4}\), \((-y^{2})^{2}=y^{4}\). Then \(15\times16x^{4}\times y^{4}=240x^{4}y^{4}\)? No, that's not matching the options. Wait, maybe the binomial is \((2x - y^{2})^{6}\), and the fifth term: let's list the terms:

- Term 1 (\(k = 0\)): \(\binom{6}{0}(2x)^{6}(-y^{2})^{0}=64x^{6}\)
- Term 2 (\(k = 1\)): \(\binom{6}{1}(2x)^{5}(-y^{2})^{1}=6\times32x^{5}\times(-y^{2})=-192x^{5}y^{2}\)
- Term 3 (\(k = 2\)): \(\binom{6}{2}(2x)^{4}(-y^{2})^{2}=15\times16x^{4}\times y^{4}=240x^{4}y^{4}\)
- Term 4 (\(k = 3\)): \(\binom{6}{3}(2x)^{3}(-y^{2})^{3}=20\times8x^{3}\times(-y^{6})=-160x^{3}y^{6}\)
- Term 5 (\(k = 4\)): \(\binom{6}{4}(2x)^{2}(-y^{2})^{4}=15\times4x^{2}\times y^{8}=60x^{2}y^{8}\)
- Term 6 (\(k = 5\)): \(\binom{6}{5}(2x)^{1}(-y^{2})^{5}=6\times2x\times(-y^{10})=-12xy^{10}\)
- Term 7 (\(k = 6\)): \(\binom{6}{6}(2x)^{0}(-y^{2})^{6}=y^{12}\)

But the problem says the first term is \(2x\) and the second term is \(-y^{2}\), which doesn't match the above. So maybe the binomial is \((2x - y^{2})^{1}\), but that's a binomial of degree 1, and the fifth term doesn't exist. There must be a misinterpretation. Wait, looking at the options for the simplified form of the fifth term, we have \(60x^{2}y^{8}\)? No, the options are \(-30x^{2}y^{8}\), \(30x^{2}y^{8}\), \(60x^{2}y^{8}\). Wait, maybe the binomial is \((2x - y^{2})^{6}\), but we made a mistake in the sign or the exponent. Wait, let's re - examine the formula for the fifth term. The options for the formula are:

1. \(\binom{6}{5}(2x)(-y^{2})^{5}\)
2. \(\binom{6}{4}(2x)^{4}(-y^{2})^{2}\)
3. \(\binom{6}{2}(2x)^{2}(-y^{2})^{4}\)

For the fifth term, the value of \(r\) (in \(T_{r + 1}\)) should be \(4\) (since \(r+1 = 5\)). So \(T_{5}=\binom{6}{4}(2x)^{6 - 4}(-y^{2})^{4}\)? No, \(6-4 = 2\), so \((2x)^{2}\), and \((-y^{2})^{4}\). But \(\binom{6}{4}=\binom{6}{2}=15\). \((2x)^{2}=4x^{2}\), \((-y^{2})^{4}=y^{8}\). Then \(15\times4x^{2}\times y^{8}=60x^{2}y^{8}\). Now, let's check the simplified form options. The options for simplified form are \(-30x^{2}y^{8}\), \(30x^{2}y^{8}\), \(60x^{2}y^{8}\). Wait, maybe there was a miscalculation. Wait, if the binomial is \((2x + y^{2})^{6}\), but the second term is \(-y^{2}\), so it's \((2x - y^{2})^{6}\). Wait, no, if we take the formula \(\binom{6}{4}(2x)^{2}(-y^{2})^{4}\), \(\binom{6}{4}=15\), \((2x)^{2}=4x^{2}\), \((-y^{2})^{4}=y^{8}\), so \(15\times4x^{2}y^{8}=60x^{2}y^{8}\). So the correct formula for the fifth term is \(\binom{6}{4}(2x)^{2}(-y^{2})^{4}\) (wait, no, the third option for the formula is \(\binom{6}{2}(2x)^{2}(-y^{2})^{4}\), and \(\binom{6}{2}=\binom{6}{4}=15\)). Then the simplified form is \(60x^{2}y^{8}\).

Step4: Simplify the formula

If the formula for the fifth term is \(\binom{6}{2}(2x)^{2}(-y^{2})^{4}\) (since \(\binom{6}{2}=\frac{6!}{2!4!}=15\)), \((2x)^{2}=4x^{2}\), \((-y^{2})^{4}=y^{8}\). Then \(15\times4x^{2}y^{8}=60x^{2}y^{8}\).

Answer:

The formula for the fifth term is \(\boldsymbol{\binom{6}{2}(2x)^{2}(-y^{2})^{4}}\) and the simplified form is \(\boldsymbol{60x^{2}y^{8}}\)