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Questions tagged [binomial-theorem]

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

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I came across this binomial sum- $p+q=99^{98}(1+99^{2})-\frac{99 \cdot 98^{98}}{1}(1+98^{2})+\frac{99 \cdot 98 \cdot 97^{98}}{1 \cdot 2}(1+97^{2})-\frac{99 \cdot 98 \cdot 97\cdot 96^{98}}{1 \cdot 2 \...
Dhairya Kumar's user avatar
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Let $x$ be a real number. Consider the following expression: $$ \sqrt[3]{\frac{x^{3} - 3x + \left(x^{2} - 1\right)\sqrt{x^{2} - 4}}{2}} + \sqrt[3]{\frac{x^{3} - 3x - \left(x^{2} - 1\right)\sqrt{x^{2} -...
TheProver's user avatar
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Using multinomial theorem the powers of terms should sum to $a + 2b + 3c + 4d + 5e = 5$; We get following solutions $e = 1$; $a = 1, d = 1$; $b = 1, c = 1$; $c = 1, a = 2$; $a = 1, b = 2$; So we get $\...
Udbhav Dikshit's user avatar
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I am reading this article by Zudilin. In page $523$, there is a formula for $a,b$ odd positive integers, $n$ a positive integer and $\tau$ a complex number, $$\sin^{2m+b}\pi n\tau=\chi_m \frac{(-1)^{(...
Max's user avatar
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The binomial formula $(x+y)^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{k}y^{n-k}$ could be interpreted, when $x,y$ are positive integers, as a two-way counting of "words" of length $n$ which use ...
digital-Ink's user avatar
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Problem statement For all $x>0$, $n, m \in \mathbb{N}$, let $$ S(x, n, m) \triangleq \sum_{j=0}^n \binom{n}{j} (1 - x)^{n-j} x^j j^3 \frac{1}{1 + \frac{j}{m} } $$ Assuming $n = o(m)$, for all $c&...
WhatAboutThat's user avatar
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Let $S_n=\sum_{r=0}^n \sum_{s=0}^n \frac{(-1)^{r+s}}{{n \choose r}\left({n \choose s}+ {n \choose r}\right)}.$ Interchange $r$ and $s$, to have $S_n=\sum_{s=0}^n \sum_{r=0}^n \frac{(-1)^{s+r}}{{n \...
Z Ahmed's user avatar
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2 votes
2 answers
148 views

Evaluate $\sum_{r=1}^{101}(-1)^{r-1}(1+\frac12+\frac13+...+\frac1r){101\choose r}$ My Attempt: If $r$ was starting from $0$, I could have written $(1-1)^{101}$. Now, maybe ${101\choose r}$ can be ...
aarbee's user avatar
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According to Mathematica, the $x \to \infty$ asymptotics of $f(x) = (x+a)^{(x+a)}$, for $a>0$, are given my $$f(x)\sim x^{(a+x)}e^{a+\frac{a^2}{2x}}.$$ How is this shown?
VerwirrterStudent's user avatar
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Prove: $$ \sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{2n - 2k}{n+1} = n 2^{n-1}. $$ Combinatorially, both sides count the number of subsets $X \subseteq A:=A_1\sqcup A_2\sqcup\ldots \sqcup A_n$ such ...
Roy Sht's user avatar
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I was just wondering if there is any specific formula for $a^{m}+b^{m}$. I have noticed that if we have $a^n+b^n$, and $n$ is odd, then we have : $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b^1+a^{n-3}b^2-...-ab^{n-...
Rishikesh Biswas's user avatar
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Suppose $0\leq a\leq 1$ and $b\geq 1, c\geq 1$ are any real numbers. Then may I know if there is any result in the literature as given below: $$\left({a+\sqrt[n]{b}}\right)^{n}\left({a+{c}}\right)\...
mathlover's user avatar
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My textbook states that: $$\sum_{i=0}^n\binom ni \left(\frac1n\right)^n$$ is equal to $$\sum_{i=0}^n\frac{n(n-1)\cdots(n-1+1)}{i!}\left(\frac1n\right)^n$$ But seeing as (n choose i) is equal to (n!/(i!...
user22960886's user avatar
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3 answers
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While working at Mathematica, I realized that $$S_m=\sum_{r=0}^{m} {n-r \choose r} x^r \tag{1}$$ is sum of two hypergeometric series when $m=n$, but when $m=[n/2]$, $S_{n/2}$ can be expressed as $$S_{...
Z Ahmed's user avatar
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Evaluate ${^5C_1}{^5C_5}-{^5C_2}{^{10}C_5}+{^5C_3}{^{15}C_5}-{^5C_4}{^{20}C_5}+{^5C_5}{^{25}C_5}$ My Attempt: We are essentially looking for the coefficient of $x^5$ in ${^5C_1}(1+x)^5-{^5C_2}(1+x)^{...
aarbee's user avatar
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3 votes
1 answer
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If $$f(x)=\sum_{r=0}^{20}(-1)^{r}{20\choose r}\left(1+3\sin x\right)^{20-r}$$ then evaluate $$\sum_{r=0}^{\infty}\left(f\left(\dfrac{\pi}{6}\right)\right)^r$$ My solution: I obtained $f(x)=\left(1+3\...
mathophile's user avatar
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4 answers
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The following question is taken from the practice set of JEE. Question: Consider $f(n)={^{2n+1}C_1}+{^{2n+1}C_3}2^3+{^{2n+1}C_5}2^6+{^{2n+1}C_7}2^9+...+{^{2n+1}C_{2n+1}}2^{3n}$ When $f(n)$ divided by $...
aarbee's user avatar
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2 votes
1 answer
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The value of ${20\choose7}-{20\choose8}+......-{20\choose20}$ is equal to (a) ${19\choose 13}$ (b) ${19\choose 14}$ (c) ${20\choose 13}$ (d) None of these My solution: $(1+x)^{20}={20\choose 0}+{20\...
mathophile's user avatar
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2 answers
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I have to prove: $$ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n} = 2^n $$ L.H.S. $$ \frac{n!}{0!(n-0)!} + \frac{n!}{1!(n-1)!} + \frac{n!}{2!(n-2)!} + \cdots + \frac{n!}{n!(n-n)!} ...
xoco's user avatar
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Just one line in Burton's number theory has made me waste hours and facepalm myself in shame. Here it is: $$\binom{n}{k} =\frac{n(n−1)···(k+1)}{(n-k)!}= \frac{n(n−1)···(n−k+1)}{k!}$$ First ...
Vince's user avatar
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In Eckle's book "Models of Quantum Matter" he gives the exercise to show that the two expressions (7.14) and (7.230) are the same (note $w_1 = (u_1,v_1),\, w_2 = (u_2, v_2)$). Could anyone ...
JvT's user avatar
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Instead of taking the angle that the revolving line makes with the x-axis to be 60 degrees, if we take it to be 75 degrees, it should converge faster using the expansion for (1 - x^2)^0.5. But, when I ...
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I want to show that $1+p$ is an element of $(\mathbb Z/p^n\mathbb Z)^\times$ and its order is $p^{n-1}$ for odd prime $p$. (Dummit/Foote 2.3.21) The textbook said one can prove it by using the ...
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Is there a fractional version of Pascal's triangle for binomial expansion similar to the standard triangle used for binomial expansion? If so, is it related to the Gamma function? \begin{array}{cccccc}...
Kendall's user avatar
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$\sum_{i=0}^{n-k} \binom{n-k}{i} i (n-1)^{n-k-1-i}$ This sum was simplified in a paper I was reading with the identitiy: $ \binom{n-k}{i} i = (n-k) \binom{n-k-1}{i-1}$ and I simply cannot find this ...
manifold97's user avatar

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