Help to solve binominal theorem

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SUMMARY

The discussion focuses on solving a problem related to the binomial theorem, specifically the expansion of the expression $$x^2\left(3x^2+\frac{k}{x}\right)^8$$ to find the constant term, which is given as 16128. The relevant term in the expansion is identified as $$3^{8-i}{8 \choose i}k^ix^{18-3i}$$. To determine the value of \( k \), the exponent of \( x \) must equal zero, leading to the equation \( 18 - 3i = 0 \), which simplifies to \( i = 6 \).

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pusekatja
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Hi!
I am struggling to solve this question about binomial theorem, could someone help me? :)

Consider the expansion of $$x^2\left(3x^2+\frac{k}{x}\right)^8$$
The constant term is 16128.
Find k.
 
Last edited:
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Hello and welcome to MHB, pusekatja! :D

Without knowing what you have tried, I can't know where you are stuck, so this is why we ask our users to show what they have tried so far.

Now, according to the binomial theorem, the $i$th term of the given expansion will be:

$$x^2{8 \choose i}\left(3x^2\right)^{8-i}\left(\frac{k}{x}\right)^i$$

How about we rewrite this term as follows:

$$3^{8-i}{8 \choose i}k^ix^{18-3i}$$

Now, what does this tell us $i$ must be in order for the term to be a constant?
 

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