find

\[\binom{100}{0} - \binom{100}{1} + \binom{100}{2} - \dots + \binom{100}{100}.\]

newsss Aug 15, 2023

#1**-1 **

This expression can be simplified using the Binomial Theorem. According to the Binomial Theorem, we have:

\[(x - y)^{100} = \binom{100}{0} x^{100} y^0 + \binom{100}{1} x^{99} y^1 + \binom{100}{2} x^{98} y^2 + \dots + \binom{100}{100} x^0 y^{100}.\]

Notice that your expression corresponds to the expansion of \((x + y)^{100}\):

\[\binom{100}{0} x^{100} y^0 - \binom{100}{1} x^{99} y^1 + \binom{100}{2} x^{98} y^2 - \dots + \binom{100}{100} x^0 y^{100}.\]

So, if we let \(x = 1\) and \(y = 1\), we obtain:

\[(1 + 1)^{100} = 2^{100}.\]

Therefore, the value of the given expression is \(2^{100}\).

SpectraSynth Aug 16, 2023

#5**0 **

The alternating sum or difference of binomial coefficients is 0, which is a fact I already knew. However, I was curious why this happens every time, so I decided to do some summation algebra to prove the result. If you are confused at any step, I can explain some steps more, but I added light commentary where I saw fit.

\(\begin{align*} \sum_{i = 0}^{n}(-1)^i \binom{n}{i} &= (-1)^0\binom{n}{0} + (-1)^n\binom{n}{n} + \sum_{i = 1}^{n - 1}(-1)^i\binom{n}{i} \\ &= 1 + (-1)^n + \sum_{i = 1}^{n - 1}(-1)^i\left[\binom{n-1}{i-1} + \binom{n - 1}{i}\right] \text{by Pascal's Rule} \\ &= 1 + (-1)^n + \sum_{i = 1}^{n - 1} (-1)^i\binom{n-1}{i-1} + (-1)^i\binom{n-1}{i} \end{align*}\)

You might wonder why I did all of this. After all, it looks like I am making the expression more complicated. However, I will use the power of a telescoping sum to simplify this monstrosity further. I apologize that I am not experienced enough with LaTeX to make this look more visually appealing.

\(\sum_{i = 1}^{n - 1} (-1)^i\binom{n-1}{i-1} + (-1)^i\binom{n-1}{i} \\ \begin{align*} i = 1:& (-1)\binom{n-1}{0} &+(-1)\binom{n-1}{1} & \\ i = 2:& &+(1)\binom{n-1}{1} & + (1)\binom{n-1}{2} \\ i = 3:&&& + (-1)\binom{n-1}{2} + (-1)\binom{n-1}{3}\\ \cdots \end{align*}\)

This makes it clear that the summation will collapse to the first and last term of the summation.

\(\begin{align*} \sum_{i = 1}^{n - 1} (-1)^i\binom{n-1}{i-1} + (-1)^i\binom{n-1}{i} &= (-1)^1\binom{n-1}{0} + (-1)^{n-1}\binom{n-1}{n-1} \\ &= -1 + (-1)^{n - 1} \end{align*}\)

Now, we put this information together and see to what the original sum evaluates.

\(\begin{align*} \sum_{i = 0}^{n}(-1)^i \binom{n}{i} &= 1 + (-1)^n -1 + (-1)^{n -1} \\ &= (-1)^n + (-1)^{n-1} \\ &= (-1)^n(1 - 1) \\ &= 0 \end{align*}\)

This concludes the proof and proves that the alternating sum or difference of binomial coefficients is 0.

The3Mathketeers Aug 16, 2023