Determine the constant term in the development of $\left(x+\frac1x\right)^6$

Question

Please help me solve this.. I think it's done with binomial theorem?

Determine the constant term in the development of $\left(x+\frac1x\right)^6$

Answer 1

The constant term is obtained with $\dbinom63 x^3\cdot\dfrac1{x^3}$. Hence we have to compute $\dbinom63$.

This can be done with Pascal's triangle. Starting with the line of coefficients for $n=4$, we deduce successively the lines for $n=5$, then $n=6$: $$\begin{matrix} &&1&&4&&6&&4&&1\\&1&&5&&10&&10&&5&&1\\ 1&&6&&15&&\color{red}{20}&&15&&6&&1 \end{matrix}$$ The constant term is $20$.

Answer 2

$(x+\frac{1}{x})^6$=$x^6+1/x^6+6 x^4+6/x^4+15 x^2+15/x^2+20$

Middle of sixth row in Pascal triangle. Odd power wouldnt have free coefficient.

Answer 3

HINT:

Apply to binomial theorem

$$\left(x+y\right)^n=\sum_{k=0}^{n}\binom{n}{k}x^ky^{n-k}$$

with $y=1/x$ and $n=6$. What value(s) of $k$ generates a constant exponent?

Answer 4

$(x+\frac1x)^6 =(x^2+\frac{1}{x^2} +2) ^3 = $

EDIT 1:

Including the term missed out before $ 6 x^2 \cdot 1/x^2 \cdot 2 =12 $

we now have $ 8 + 12 =20. $

Answer 5

Use the binomial theorem: $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k}b^k.$$ In your case, $n=6$, $a=x$ and $b=1/x$. Since you need the constant term: $$C = \binom{6}{k} x^{6-k}\left(\frac{1}{x}\right)^k = \binom{6}{k} \frac{x^{6-k}}{x^k} = \binom{6}{k} x^{6-2k} \iff k=...$$

Answer 6

when will there be a constant term? when the $x^p$ and $\frac {1}{x^q}$ cancel out each other isn't it? so that would happpen when $p=q$ and since powers from a binomial expansion so $p+q=6$ this implies $2p=6$ i.e $p=3$.Thus the constant term is $6\choose 3$ $=20$.This is your constant term

Answer 7

You can do that without knowing the theory.

$$(x+x^{-1})(x+x^{-1})=x^2+1+1+x^{-2}=x^2+2+x^{-2}$$ $$(x^2+2+x^{-2})(x+x^{-1})=x^3+2x+x^{-1}+x+2x^{-1}+x^{-3}=x^3+3x+3x^{-1}+x^{-3}\\ \cdots.$$

Just writing the coefficients, you reconstitute the Pascal pattern

$$1\ 1\\1\ 2\ 1\\1\ 3\ 3\ 1\\1\ 4\ 6\ 4\ 1\\1\ 5\ 10\ 10\ 5\ 1\\1\ 6\ 15\ \color{green}{20}\ 15\ 6\ 1$$

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Combinatorial reasoning:

Without regrouping, the development of $(x+x^{-1})^6$ is made of terms which are the product of factors $x$ and $x^{-1}$, such as $xx^{-1}xx^{-1}x^{-1}x^{-1}=x^{-2}$, with a total of $6$ factors $x$ or $x^{-1}$.

You are looking for the constant terms, so you need to consider all sums of $6$ exponents $\pm1$ giving $0$, i.e. all the ways to shuffle $+++---$. There are $6$ exponents, which you can permute in $6!=720$ ways. But the $3+$ and $3-$ are undistinguishable, so that every permutation is repeated $3!\cdot3!=36$ times. Hence there are

$$\frac{6!}{3!\cdot3!}=20$$ combinations corresponding to $20$ terms $x^0$.