# Question Video: Using Binomial Coefficients to Solve Problems Mathematics

Answer the following questions for the expansion of (2 + 𝑘𝑥)⁶. Given that the coefficient of 𝑥² = 60, and 𝑘 is positive, find 𝑘. Hence, using your value of 𝑘, work out the coefficient of 𝑥⁵ in the expansion.

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### Video Transcript

Answer the following questions for the expansion of two plus 𝑘𝑥 to the sixth power. Given that the coefficient of 𝑥 squared is 60 and 𝑘 is positive, find 𝑘. Hence, using your value of 𝑘, work out the coefficient of 𝑥 to the fifth power in the expansion.

We will answer this question using the binomial theorem. The expression two plus 𝑘𝑥 to the sixth power is written in the form 𝑎 plus 𝑏 to the 𝑛th power. The binomial theorem states that this is equal to 𝑎 to the power of 𝑛 plus 𝑛𝐶 one multiplied by 𝑎 to the power of 𝑛 minus one multiplied by 𝑏 plus 𝑛𝐶 two multiplied by 𝑎 to the power of 𝑛 minus two multiplied by 𝑏 squared and so on up to 𝑏 to the power of 𝑛, where 𝑛 must be an integer value greater than zero.

In this question, 𝑎 is equal to two, 𝑏 is equal to 𝑘𝑥, and the exponent 𝑛 is equal to six. Whilst we could write out the full expansion, the first part of our question asks us about the coefficient of 𝑥 squared. This will be the third term of our expansion as it will be the one containing 𝑏 squared. This term is equal to six 𝐶 two multiplied by two to the power of four multiplied by 𝑘𝑥 squared. We know that 𝑛𝐶𝑟 is equal to 𝑛 factorial divided by 𝑟 factorial multiplied by 𝑛 minus 𝑟 factorial. This means that six 𝐶 two is equal to six factorial divided by two factorial multiplied by four factorial. Six factorial can be rewritten as six multiplied by five multiplied by four factorial. Our expression simplifies to 30 divided by two, which is equal to 15.

Two to the fourth power is equal to 16, and 𝑘𝑥 all squared is equal to 𝑘 squared 𝑥 squared. The third term in our expansion simplifies to 240𝑘 squared 𝑥 squared. We are told that the coefficient of 𝑥 squared is 60. This means that 240𝑘 squared equals 60. Dividing through by 240, we have 𝑘 squared equals one-quarter. We can then square root both sides of this equation. And noting that 𝑘 must be positive, the value of 𝑘 is one-half. If the coefficient of 𝑥 squared in the expansion two plus 𝑘𝑥 to the sixth power is equal to 60 and 𝑘 is positive, then the value of 𝑘 is one-half.

We will now clear some space and consider the second part of the question. As 𝑘 is equal to one-half, we can rewrite our expression as two plus a half 𝑥 raised to the sixth power. We are now interested in the term in the expansion containing 𝑥 to the fifth power. This is equal to six 𝐶 five multiplied by two to the first power multiplied by a half 𝑥 to the fifth power. We know that 𝑛𝐶 𝑛 minus one is simply equal to 𝑛. So, in our expression, six 𝐶 five equals six. Two to the first power is simply two, and a half 𝑥 to the fifth power is one over 32 𝑥 to the fifth power. This simplifies to 12 over 32 𝑥 to the fifth power. And as both 12 and 32 are divisible by four, the fraction simplifies to three-eighths. We have been asked to work out the coefficient of 𝑥 to the fifth power in the expansion, and this is therefore equal to three-eighths.

The two answers to the question are 𝑘 equals one-half and three-eighths.

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