Answer:
k = 2
Step-by-step explanation:
Binomial Theorem
[tex]\displaystyle (a+b)^n=\binom{n}{0}a^{n-0}b^0+\binom{n}{1}a^{n-1}b^1+...+\binom{n}{r}a^{n-r}b^r+...+\binom{n}{n}a^{n-n}b^n[/tex]
[tex]\textsf{Where }\displaystyle \rm \binom{n}{r} \: = \:^{n}C_{r} = \frac{n!}{r!(n-r)!}[/tex]
Factorial is denoted by an exclamation mark "!" placed after the number. It means to multiply all whole numbers from the given number down to 1.
Example: 4! = 4 × 3 × 2 × 1
Given expression:
[tex]x^7\left(\dfrac{x^4}{2}+\dfrac{k}{x^3}\right)^7[/tex]
Therefore:
[tex]\implies x^7\left(\dfrac{x^4}{2}+\dfrac{k}{x^3}\right)^7= x^7(a+b)^n[/tex]
[tex]\implies a=\dfrac{x^4}{2}[/tex]
[tex]\implies b=\dfrac{k}{x^3}[/tex]
[tex]\implies n=7[/tex]
To find the value of the constant, we need to find the value of r for the term in x⁰ since x⁰ = 1. Therefore, ignoring the coefficient:
[tex]\implies x^7\left(a^{7-r} \cdot b^r\right)=x^{0}[/tex]
[tex]\implies \dfrac{x^7\left(a^{7-r} \cdot b^r\right)}{x^7}=\dfrac{x^{0}}{x^7}[/tex]
[tex]\implies \left(a^{7-r} \cdot b^r\right)=x^{-7}[/tex]
[tex]\implies \left(\left(\dfrac{x^4}{2}\right)^{7-r} \cdot \left(\dfrac{k}{x^3}\right)^r\right)=x^{-7}[/tex]
[tex]\implies \dfrac{x^{28-4r}}{2^{7-r}} \cdot \dfrac{k^r}{x^{3r}}=x^{-7}[/tex]
Ignore the constants:
[tex]\implies x^{28-4r} \cdot x^{-3r}=x^{-7}[/tex]
[tex]\implies x^{28-7r}=x^{-7}[/tex]
[tex]\implies 28-7r=-7[/tex]
[tex]\implies -7r=-35[/tex]
[tex]\implies r=5[/tex]
Therefore, the term in x⁰ will be the 6th term, where r = 5. Substitute r = 5 into the r term of the binomial expansion, set it to 168 and solve for k:
[tex]\implies \displaystyle x^7\left(\binom{7}{5}a^{7-5}b^5\right)=168[/tex]
[tex]\implies x^7 \left( \dfrac{7!}{5!(7-5)!}a^{2}b^5\right)=168[/tex]
[tex]\implies x^7 \left(\dfrac{7!}{5! \:2!}a^{2}b^5\right)=168[/tex]
[tex]\implies x^7\left(\dfrac{7 \times 6 \times \diagup\!\!\!\!\!5 \times \diagup\!\!\!\!\!4 \times \diagup\!\!\!\!\!3 \times \diagup\!\!\!\!\!2 \times \diagup\!\!\!\!\!1}{\diagup\!\!\!\!\!5 \times \diagup\!\!\!\!\!4 \times \diagup\!\!\!\!\!3 \times \diagup\!\!\!\!\!2 \times \diagup\!\!\!\!\!1 \times 2 \times 1}a^2b^5\right)=168[/tex]
[tex]\implies x^7 \left(\dfrac{42}{2}a^2b^5\right)=168[/tex]
[tex]\implies x^7 \left(21a^2b^5\right)=168[/tex]
[tex]\implies x^7 \left(21\left(\dfrac{x^4}{2}\right)^2\left(\dfrac{k}{x^3}\right)^5\right)=168[/tex]
We can ignore the variables in x at this point and just concentrate on the constants, but I have left them in the calculation to prove they are cancelled out.
[tex]\implies x^7 \left(21\left(\dfrac{x^8}{4}\right)\left(\dfrac{k^5}{x^{15}}\right)\right)=168[/tex]
[tex]\implies x^7\left( \dfrac{21k^5x^8}{4x^{15}}\right)=168[/tex]
[tex]\implies \dfrac{21k^5x^8x^7}{4x^{15}}=168[/tex]
[tex]\implies \dfrac{21k^5x^{15}}{4x^{15}}=168[/tex]
[tex]\implies \dfrac{21k^5}{4}=168[/tex]
[tex]\implies k^5=32[/tex]
[tex]\implies k=\sqrt[5]{32}[/tex]
[tex]\implies k=2[/tex]
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