Using binomial theorem expand
[tex]( x - \frac{1}{2 {x}^{2} } ) {}^{ 4} [/tex]
and represent its coefficients using Pascal's Triangle.
please answer correctly.. I have to submit my assignment today, so please.
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Using binomial theorem expand
[tex]( x - \frac{1}{2 {x}^{2} } ) {}^{ 4} [/tex]
and represent its coefficients using Pascal's Triangle.
please answer correctly.. I have to submit my assignment today, so please.
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Answer:
To expand \((x - \frac{1}{2}x^2)^4\) using the binomial theorem, we can use the formula:
\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]
In this case, \(a = x\) and \(b = -\frac{1}{2}x^2\), and \(n = 4\).
The expansion will be:
\[\binom{4}{0}x^4 \left(-\frac{1}{2}x^2\right)^0 + \binom{4}{1}x^3 \left(-\frac{1}{2}x^2\right)^1 + \binom{4}{2}x^2 \left(-\frac{1}{2}x^2\right)^2 + \binom{4}{3}x \left(-\frac{1}{2}x^2\right)^3 + \binom{4}{4} \left(-\frac{1}{2}x^2\right)^4\]
Now, let's simplify this expression:
\[\binom{4}{0}x^4 - \binom{4}{1}\frac{1}{2}x^5 + \binom{4}{2}\frac{1}{4}x^6 - \binom{4}{3}\frac{1}{8}x^7 + \binom{4}{4}\frac{1}{16}x^8\]
Now, let's calculate the coefficients using Pascal's Triangle:
\[\binom{4}{0} = 1, \quad \binom{4}{1} = 4, \quad \binom{4}{2} = 6, \quad \binom{4}{3} = 4, \quad \binom{4}{4} = 1\]
Substitute these coefficients back into the expression:
\[x^4 - 2x^5 + \frac{3}{2}x^6 - \frac{1}{2}x^7 + \frac{1}{16}x^8\]
So, the expanded form of \((x - \frac{1}{2}x^2)^4\) is \(x^4 - 2x^5 + \frac{3}{2}x^6 - \frac{1}{2}x^7 + \frac{1}{16}x^8\), and the coefficients were derived using Pascal's Triangle.
Step-by-step explanation:
Answer:
I tried my best. let me know if this helps you or not.. mark me as brainliest please :)
Step-by-step explanation:
Using the binomial theorem, we can expand [tex]( x - \frac{1}{2 {x}^{2} } ) {}^{ 4}[/tex] as follows:
[tex]( x - \frac{1}{2 {x}^{2} } ) {}^{ 4} = \sum_{k=0}^{4} {4 \choose k} x^{4-k} \left( \frac{-1}{2 {x}^{2}} \right)^{k}[/tex]
where [tex]{4 \choose k}[/tex] are the binomial coefficients. We can represent these coefficients using Pascal's Triangle as follows:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
The first row corresponds to [tex]{4 \choose 0}[/tex], the second row corresponds to [tex]{4 \choose 1}[/tex], and so on. Therefore, we can rewrite the expanded form of the expression as:
[tex]( x - \frac{1}{2 {x}^{2} } ) {}^{ 4} = {4 \choose 0} x^{4} \left( \frac{-1}{2 {x}^{2}} \right)^{0} + {4 \choose 1} x^{3} \left( \frac{-1}{2 {x}^{2}} \right)^{1} + {4 \choose 2} x^{2} \left( \frac{-1}{2 {x}^{2}} \right)^{2} + {4 \choose 3} x \left( \frac{-1}{2 {x}^{2}} \right)^{3} + {4 \choose 4} \left( \frac{-1}{2 {x}^{2}} \right)^{4}[/tex]
Simplifying this expression gives:
[tex]( x - \frac{1}{2 {x}^{2} } ) {}^{ 4} = x^{4} - 2 x^{2} + \frac{3}{16 {x}^{4}}[/tex]
Therefore, the coefficients of the expanded expression are:
- The coefficient of [tex]x^{4}[/tex] is [tex]{4 \choose 0} = 1[/tex].
- The coefficient of [tex]x^{3}[/tex] is [tex]{4 \choose 1} \cdot \frac{-1}{2} = -2[/tex].
- The coefficient of [tex]x^{2}[/tex] is [tex]{4 \choose 2} \cdot \frac{1}{4} = 3[/tex].
- The coefficient of [tex]x^{1}[/tex] is [tex]{4 \choose 3} \cdot \frac{-1}{8} = 0[/tex].
- The coefficient of [tex]x^{0}[/tex] is [tex]{4 \choose 4} \cdot \frac{1}{16} = \frac{3}{16}[/tex].
I hope this helps!