+0

# help

0
49
5
+52

find

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

Aug 15, 2023

#1
+121
-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 $$2^{100}$$.

Aug 16, 2023
#3
0

Deleted ...............

Guest Aug 16, 2023
edited by Guest  Aug 16, 2023
edited by Guest  Aug 16, 2023
#4
0

∑[ (-1)^n * bin(100, n), n,  0,  100 ]==0

Aug 16, 2023
#5
+177
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.

Aug 16, 2023