Expansion problem!!

Algebra Level 5

If the last term in the binomial expansion of $\left( 2^{1/3}-\frac{1}{\sqrt{2}} \right)^{n}$ is $\left( \frac{1}{3^{\frac{5}{3}}} \right)^{\log_{3}{8}}$ , then what will the 5th term be?

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The answer is 210.

2 solutions

Jesus Ulises Avelar
Feb 24, 2015

I solve this using the binomial theorem .

Same and exactly the same method as I used. Cheers!

Kishore S. Shenoy - 5 years, 9 months ago

Jake Lai
Feb 24, 2015

First, we simplify.

$\left( \frac{1}{3^{\frac{5}{3}}} \right)^{\log_{3} 8} = 3^{-5\log_{3} 8/3} = 2^{-5} = (\frac{1}{\sqrt{2}})^{10}$

This implies $n = 10$ . Now, by the binomial theorem, the 5th term will be

${10 \choose 6}(2^{1/3})^{6}(-\frac{1}{\sqrt{2}})^{10-6} = {10 \choose 6} = \boxed{210}$

Doesn't the question ask for the 5th term however? Also isn't the term you calculated above, the 6th term not the 7th term?

Sadat Shaik - 6 years, 3 months ago

It's the 7th term.

Because in the binomial theorem $\displaystyle (a+b)^{n} \sum_{k=0}^{n} {n \choose k}a^{k}b^{n-k}$ you are computing a total of 11 terms; since the convention is the expression above (even though you could switch the indices $k$ and $n-k$ if you wanted to) the 7th term is then at $k=6$ .

The author probably made a typo, since the terms at $k = 4,5$ do not give an integral value (the input is integral). I've disputed the problem.

Edit: See Calvin's comment.

Jake Lai - 6 years, 3 months ago

Jake, you have your order of terms mixed up. The first term of $(a+b)^n$ is $a ^ n$ , which is not obtained by setting $k = 0$ in your expression.

Instead, it should be $( a+b)^n = \sum_{k=0}^n { n \choose k } a ^ { n-k } b^{k}$ .

This explains the discrepancy between your "7th term" and his "5th term".

Calvin Lin Staff - 6 years, 3 months ago

@Calvin Lin – Argh, yup. I've made the necessary change to the problem and my solution. Thanks.

Jake Lai - 6 years, 3 months ago

Expansion problem!!

The answer is 210.

2 solutions

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