Finding the coefficient of $x^7$ in the expansion $(1 + x)^{23}$

Question

By definition, the Binomial Theorem states:

$$(x+y)^n = {n\choose 0}x^n + {n\choose 1}x^{n-1}y + {n\choose 2}x^{n-2}y^2 + \cdots + {n \choose {n-1}}xy^{n-1} + {n \choose n}y^n$$

For any $x,y\in\mathbb R$ and $n \in \mathbb N$.

Now I know I could expand this simply by using the binomial theorem as stated above and just look at the coefficient of $x$ where $n = 7$, however that gets very lengthy, especially when - like in this case - our $n$ is a large number.

I was hoping for a better understanding through an explanation of how I may go about tackling this problem without expanding to the $n$th term.

If I set y = 1 (like I have in the title), at which term from the expansion am I curious about? — taylor.tackett, Jun 14 '16 at 06:16
Then you have $(1+x)^{23}=1+23x+\dots+{23\choose7}x^7+\dots+x^{23}$. — almagest, Jun 14 '16 at 06:17
It will be the one with 7 at the bottom. It can help to know that n choose k is symmetric. — mathreadler, Jun 14 '16 at 06:18
Oh, sorry, I see your difficulty. Put $x=1$ in the binomial expansion and you have $(1+y)^n=1+{n\choose1}y+{n\choose2}y^2+\dots+{n\choose k}y^k+\dots+y^n$. Alternatively remember that ${n\choose n-k}={n\choose k}$. — almagest, Jun 14 '16 at 06:20

score 5 · Accepted Answer · answered Jun 14 '16 at 06:20

5

In the expansion of $(a+x)^n$, the general term, that is the $(r+1)^{th}$ term is $$T_{r+1}=\binom nr a^{n-r}x^r.$$

Thus, in the expansion of $(1+x)^{23}$, the general term is $$\binom{23}r1^{23-r}x^r.$$

You want the coefficient of $x^7$, that is $r=7$, so the the coefficient is $$\binom{23}7.$$

This is the way, using general terms, you can derive any coefficient of any term without any expansion.

answered Jun 14 '16 at 06:20

That is exactly what I was looking for; a clear and simply form. Thank you. – taylor.tackett Jun 14 '16 at 06:26

score 3 · Answer 2 · 2016-06-14T09:41:29.487

3

The coefficient of $x^7$ will be

$$\binom{23}7=\frac{23!}{7!16!}.$$

An efficient way to compute that ratio is by incrementally evaluating $\binom{23}k$ for increasing $k$, as these are simply related to each other and are integer.

$$\binom{23}0=1,\\ \binom{23}1=1\frac{23}1=23,\\ \binom{23}2=23\cdot\frac{22}2=23\cdot11,\\ \binom{23}3=23\cdot11\cdot\frac{21}3=23\cdot11\cdot7,\\ \binom{23}4=23\cdot11\cdot7\cdot\frac{20}{4}=23\cdot11\cdot7\cdot5,\\ \binom{23}5=23\cdot11\cdot7\cdot5\cdot\frac{19}{5}=23\cdot11\cdot7\cdot19,\\ \binom{23}6=23\cdot11\cdot7\cdot19\cdot\frac{18}{6}=23\cdot11\cdot7\cdot19\cdot3,\\ \binom{23}7=23\cdot11\cdot7\cdot19\cdot3\cdot\frac{17}{7}=23\cdot11\cdot19\cdot3\cdot17.$$

Hence

$$245157\,x^7.$$

(You can also write the long fraction and simplify.)

edited Jun 14 '16 at 09:41

answered Jun 14 '16 at 08:58

+1 Because I find it depressing that a statement of something that the OP and others can memorise for mindless application attracts 6 votes plus a green tick, whilst a useful explanation of some theory only attracts 1! – almagest Jun 14 '16 at 12:00
@almagest: they call it democracy :) – Jun 14 '16 at 12:01
@almagest: democracy strikes back, it seems... – Jun 14 '16 at 13:15
@SubhadeepDey My comment was not meant as a criticism of you. I agree with you, if that is what the OP wants, then someone should provide it. My comment simply an expression of my depression that in a site where most of the questions are asked by undergraduates so many of them should upvote this kind of answer. – almagest Jun 14 '16 at 13:18

score 1 · Answer 3 · answered Jun 14 '16 at 12:07

It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.

This way we can write \begin{align*} [x^7](1+x)^{23}&=[x^7]\sum_{n=0}^{23}\binom{23}{n}x^n\tag{1}\\ &=\binom{23}{7}\tag{2}\\ &=\frac{23!}{7!16!}\\ \end{align*}

Comment:

In (1) we use the binomial series expansion
In (2) we select the coefficient of $x^7$

Finding the coefficient of $x^7$ in the expansion $(1 + x)^{23}$

3 Answers3

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