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Question Video: Finding the Exponent of a Binomial given the Order of the Term Free of Variables Mathematics

If π‘Žβ‚β‚‚ is the term free of π‘ž in (6π‘ž βˆ’ (1/π‘žΒ²))^(𝑛), find 𝑛.

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Video Transcript

If π‘Ž sub 12 is the term free of π‘ž in six π‘ž minus one over π‘ž squared to the power of 𝑛, find 𝑛.

We recall that the general term in any binomial expansion of the form π‘Ž plus 𝑏 to the 𝑛th power is written π‘Ž sub π‘Ÿ plus one. This is equal to 𝑛 choose π‘Ÿ multiplied by π‘Ž to the power of 𝑛 minus π‘Ÿ multiplied by 𝑏 to the power of π‘Ÿ. In this question, π‘Ž is equal to six π‘ž and 𝑏 is equal to negative one over π‘ž squared. The exponent or power is 𝑛. And π‘Ÿ is equal to 11, as we are dealing with the 12th term. The value of π‘Ÿ is one less than this.

Substituting in these values, we see that the 12th term is equal to 𝑛 choose 11 multiplied by six π‘ž to the power of 𝑛 minus 11 multiplied by negative one over π‘ž squared to the power of 11. Focusing on the last part, we can raise both the numerator and denominator to the 11th power, giving us negative one to the power of 11 over π‘ž squared to the power of 11. This is equal to negative one over π‘ž to the power of 22.

In order to simplify π‘ž squared raised to the power of 11, we multiply the exponents. And two multiplied by 11 is 22. As we are looking for the term that is free of π‘ž, we know that the exponent of six π‘ž must be equal to 22, as this would give us six π‘ž to the power of 22 multiplied by negative one over π‘ž to the power of 22. This can be rewritten as six to the power of 22 multiplied by π‘ž to the power of 22 multiplied by negative one over π‘ž to the power of 22. As π‘ž cannot be equal to zero, these cancel, which just leaves us with six to the power of 22 multiplied by negative one.

In order to find the value of 𝑛, we need to solve 𝑛 minus 11 is equal to 22. We can add 11 to both sides of this equation so that 𝑛 is equal to 33. If π‘Ž sub 12 is the term free of π‘ž in six π‘ž minus one over π‘ž squared to the power of 𝑛, then 𝑛 is equal to 33.

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