# Questions tagged [binomial-theorem]

For questions related to the binomial theorem, which describes the algebraic expansion of powers of binomials.

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### How does this vandermonde identity proof works?

Recently I have taken a look at this vandermonde identity proof Inductive Proof for Vandermonde's Identity?. \begin{align*} \binom{m + (n+1)}r &= \binom{m+n}r + \binom{m+n}{r-1}\\ &...
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### binomial coefficients with the given function

A sequence is given as $f(n)=\dfrac{1+dx+ex^2}{1+ax+bx^2+cx^3}$.we have to calculate value of $f(n)$.we know the value of $a,b,c,d,e,n$. and we have to find $f(n)$. I think the answer should be from ...
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### How do I prove that ${n-1 \choose k} - {n-1 \choose k-1}$ is equal to ${n-2k\over n}{n \choose k}$? [closed]

I really need help as I am really struggling here on what to do. $${n-1\choose k}-{n-1\choose k-1}=\frac{n-2k}n{n\choose k}$$
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### Why does $(6+i)^3 = 198+107i$?

When I expand it I get $i^3+18i^2+108i+216$. How does one go from that to $198+107i$. I noticed that the 1st term of the 2nd equation is = to the 4th term of the 1st equation - the 2nd term of the 1st ...
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### Trace of $(A+B)^n$ with B an involutory matrix

Consider two matrices $A$ and $B$ that do not commute, so that the binomial theorem does not apply. However, one of them (say $B$) is an involutory matrix, meaning that $B^2 = I$. I am wondering ...
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### Combinatorial Summation

I'm trying to solve the following question: If $s_n$ is the sum of first $n$ natural numbers, then prove that $$2(s_1s_{2n}+s_2s_{2n-1}+\dots+s_ns_{n+1})=\frac{(2n+4)!}{5!(2n-1)!}$$ This is where I'...
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### Coefficient of $x^{12}$ in $(1+x^2+x^4+x^6)^n$

I need to find the coefficient of $x^{12}$ in the polynomial $(1+x^2+x^4+x^6)^n$. I have reduced the polynomial to $\left(\frac{1-x^8}{1-x^2}\right)^ n$ and tried binomial expansion and Taylor series, ...
I have been given this question: "Find the value of $c$ if, in the expansion of $(cx + 2)^3$, the coefficient of $x$ is $24$" To solve this question, I have tried using the 'General Term In ...