Suppose $𝑋 \sim 𝐺𝑒𝑜𝑚(𝑝_1)$ on $\{1, 2, 3, \dots\}$, $𝑌 \sim 𝐺𝑒𝑜𝑚(𝑝_2)$ on $\{1, 2, 3, \dots \}$, and 𝑋 ⊥ 𝑌. Let $𝑆 = 𝑋 + 𝑌$. Find $𝑃(𝑆 = 𝑠)$ under each condition.   a) Assume $𝑝 = 𝑝_1 = 𝑝_2$.  b) Assume $𝑝_1 \neq 𝑝_2$. After setting up your summation, identify $𝑎, 𝑟$ and $𝑛$ in a finite geometric series using the below formula and evaluate your summation.  $$𝑎 + 𝑎𝑟 + ⋯ + 𝑎𝑟^{(𝑛−1)} = \frac{𝑎(1 − 𝑟^𝑛)}{ 1 - r}$$  EDIT: So part a) so far I know I should use negative binomial distribution for part a (# of trials until 2 successes). So I could use the formula $\binom{x-1}{r-1}\cdot  p ^ r \cdot (1 - p)^{(x - r)}$ with $r = 2$. But I don't know how to continue from that.
120996
If $X$ and $Y$ are independent geometric random variables, $X \sim G(p)$ and $Y \sim G(q)$ then if $Z = X+Y$ I need to show that:     $$P[Z=z] = \frac{pq}{p-q}[(1-q)^{z-1}-(1-p)^{z-1}]$$   My attempt: \begin{align} P[Z=z] &amp;= \sum_{k=1}^z P[X=k]P[Y=z-k] &amp;&amp; \text{(not sure of my summation limits here)}  \\ &amp;= \sum_{k=1}^z p(1-p)^{k-1} \cdot q(1-q)^{z-k-1} \\ &amp;= \frac{pq}{(1-p)}\cdot \frac{(1-q)^z}{(1-q)}\cdot\sum_{k=1}^z (\frac{1-p}{1-q})^k \\ &amp;= \frac{pq}{(1-p)}\cdot \frac{(1-q)^z}{(1-q)} \cdot \frac{1-(\frac{1-p}{1-q})^k}{1-\frac{1-p}{1-q}} \\ &amp;= \frac{pq}{(1-p)}\cdot \frac{(1-q)^z}{(p-q)} \left[1 -(\frac{1-p}{1-q})^k\right] \\ &amp;= \frac{pq}{p-q}\left[\frac{(1-q)^z}{1-p} - \frac{(1-p)^z}{1-p}\right] \end{align}  The result is pretty close to the answer. The only discrepancy is the first expression in the parentheses, $\frac{(1-q)^z}{1-p}$ which should be $\frac{(1-q)^z}{1-q}$.    Could someone please have a look at my working and show me where I have gone wrong? Or perhaps a neater calculation? Thanks!    PS I did check similar questions answered here, but couldn't find anything relevant to my problem.
108190
Question :   Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$   What I have done :   nth term of numerator and denominator is $2r-1$ and $r(r+1)(r+2)$ respectively.  Therefore the nth term of given series is :   $\frac{2r-1}{r(r+1)(r+2)} =\frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2}$ .....(1)  By using partial fraction :   and solving for A,B and C we get A = 1/2, B = -1, C =1/2  Putting the values of A,B and C in (1) we get :   $\frac{1}{2r}-\frac{1}{r+1}+\frac{1}{2(r+2)}$  But by putting $r =1,2,3, \cdots$  I am not getting the answer. Please guide how to solve this problem . Thanks.
33804
$$\sum_{r=1}^nr(r+1)=\frac{n(n+1)(n+2)}{3}$$  could you help me with how exactly I work this out?
68698
Do you have any idea to prove the following identity via a combinatorial (or algebraic) method?  $\sum_{n=r}^{n=\infty} {n \choose r}^{-1} = \frac{r}{r-1}$  This is  of the book Chen C.C., Koh K.M. Principles and techniques in combinatorics. The book does not give a solution, although it mentions: "see H. W. Gould, Combinatorial Identities, Morgantown, W.V. (1972), 18-19".  many thanks in advance, Shahram
18019
I'm trying to prove that ${n \choose r}$ is equal to ${{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$.  I suppose I could use the counting rules in probability, perhaps combination= ${{n} \choose {r}}=\frac{n!}{r!(n-r!)}$.  I want to see an actual proof behind this equation.  Does anyone have any ideas?
22081
I want to prove the following: $$\sum_{k=1}^{n}k\binom{n}{k}=n\cdot2^{n-1}$$ what I did is(use binominal): $$\binom{n}{k}X^k\cdot 1^{n-k} = (X+1)^n$$ $$k\binom{n}{k}X^k\cdot 1^{n-k} = k(X+1)^k-1$$ now I replace $k$ by $n$ and insert $X=1$ , I will get now $n\cdot2^{n-1}$ what you are suggesting? there is another way to do that? thanks!
23973
Q: For $p\in\mathbb{R}$ such that $2&lt;p&lt;\infty$, and $a,b&gt;0$, prove the following  $$a^p+b^p&lt;(a^2+b^2)^{p/2}$$  My attempt: I say it's equivalent to proving that $(a^p+b^p)^{1/p}&lt;(a^2+b^2)^{1/2}$. Then I did binomial expansions on both sides of the inequality. Both the expansions have $a+b$ so I canceled them. For the rest of the terms, I thought I'd get geometric series on both sides and show that the series on the left is smaller than that of the right by using related sum formulas. But what I found were not quite geometric. For the left side  $$\frac{1}{p}a^{1-p}b^p+\frac{1-p}{2p^2}a^{1-2p}b^{2p}+\frac{(1-p)(1-2p)}{6p^3}a^{1-3p}b^{3p}+\ldots$$  While for the right  $$\frac{1}{2}a^{-1}b^2-\frac{1}{8}a^{-3}b^4+\frac{1}{16}a^{-5}b^6+\ldots$$  How to proceed to prove the inequality?
4684
First we know the fact that $p$ divides $\binom {p}{k}$ for $k\in \{1,\ldots,p-1\}$.  Second we have : $\binom {p+k}{k}= \frac{(p+k)\cdots(p+1)}{k!}$ which means : $k!\binom {p+k}{k}=(p+k)\cdots(p+1)$.  Notice that every term of the product is divided by $p$ except the last product $k(k-1)\cdots1=k!$  So $p$ divides $k!(\binom {p+k}{k} - 1)$. By Gauss lemma for $k\in \{1,\ldots,p-1\}$, $p$ divides $(\binom {p+k}{k} - 1)$.  Then $p^2$ divides $\binom{p}{k}(\binom {p+k}{k} - 1)$. We obtain : $\binom{p}{k}(\binom {p+k}{k} \equiv \binom{p}{k} \pmod{p^2}$.  So : $\sum \limits_{k=0}^{p} \binom {p}{k} \binom{p+k}{k}\equiv 1 +\binom{2p}{p} + \sum \limits_{k=1}^{p-1} \binom{p}{k} \equiv 1 +\binom{2p}{p} +2^{p}-2 \pmod{p^2}$  Then I'm stuck...  Maybe it is possible to prove that $\binom{2p}{p} \equiv 2\pmod{p^2}$..  Thanks in advance !
41783
Whilst working out  the following summation emerged: $$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$ The is equivalent to  $$\begin{align} \sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&amp;=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}+\cdots +\frac{1\cdot 3\cdot 5\cdot \cdots \cdot(2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot 2n}\\ &amp;=\frac 12\left(1+\frac 34\left(1+\frac 56\left(1+\cdots \left(1+\frac {2n-1}{2n}\right)\right)\right)\right) \end{align}$$  and terms are the same as coefficients in the expansion of $(1-x)^{-1/2}$.  Once the solution  $$ \frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}$$ is known, the telescoping sum can be easily derived, i.e.    $$\frac 1{2^{2m}}\binom {2m}m=\frac {m+1}{2^{2(m+1)-1}}\binom {2(m+1)}{m+1}-\frac m{2^{2m-1}}\binom {2m}m$$  However, without knowing this a priori, how would we have approached this problem?
22985
The series is given by ($a&gt;0$) $$\frac{1}{a} +\frac{1}{a+1} - \frac{1}{a+2} +\frac{1}{a+3} +\frac{1}{a+4} -\frac{1}{a+5} +\cdots$$ So how can I move forward? I  can not find the general term  of the series. If someone can give the general term then I can go further.
47437
Use induction to prove that    $\displaystyle\sum_{r=1}^n r\cdot r! =(n+1)! -1$   I first showed that the formula holds true for $n=1$.  Then I put n as $k$ and got an expression for the sum in terms of $k$.  I then found the sum till the $(k+1)$th term by adding the $(k+1)$th term to both sides of the equation and compared it to the sum expression I get by plugging $k+1$ into the sum expression given in the question - they don't match.
120952
Suppose $X=[X_{1},X_{2}]$ and $X$~$N_2(μ,Σ)$. I wish to find the distribution of  $5X_{1}^2+2X_{1}X_{2}+X_{2}^2$. Since this is of a quadratic form I do not know a way of solving this. However I kind of feel like its chi-squared distributed with parameter 2. Any answers will be welcomed. Thanks
68303
This is a 2 part problem.  Part I  I need help finding a formula for this sequence of numbers:  $$\frac{1} {1\times 2} + \frac {1} {2\times3} + \cdots + \frac {1} {n(n+1)}$$  Part II  I need to prove the formula conjectured in Part I.
9576
We know that the formula for computing a geometric series is:$$\sum_{i=1}^{\infty}{a_0r^{i-1}} = \frac{a_0}{1-r}$$ Out of curiosity, I would like ask: Is there any ways the formula can be derived other than the following two ways?  Method 1 (The way I found on my own): $$\sum_{i=1}^{\infty}{a_0r^{i-1}} \equiv S$$ $$S = a_0r^0+a_0r^1+a_0r^2+\cdots$$ $$S = r\left(a_0r^{-1} + a_0r^{0} + a_0r^1+\cdots\right)$$ $$S = r\left(a_0r^{-1} + S\right)$$ $$S = a_0 + rS$$ $$(1-r)S = a_0$$ $$S = \frac{a_0}{(1-r)}$$ Note that for this to work, you must first confirm this: $$\lim_{n\to\infty} a_n = 0$$ Method 2 (The way I found on the web): $$\sum_{i=1}^{n}{a_0r^{i-1}} \equiv S_n$$ $$S_n = a_0r^0+a_0r^1+a_0r^2+\cdots + a_0 r^{n-2} + a_0 r^{n-1}$$ $$rS_n = r\left(a_0r^0+a_0r^1+a_0r^2+\cdots + a_0 r^{n-2} + a_0 r^{n-1}\right)$$ $$rS_n = a_0r^1 + a_0r^2 + a_0r^3 + \cdots + a_0 r^{n-1} + a_0 r^{n}$$ $$S_n-rS_n = a_0r^0 - a_0r^n$$ $$(1-r)S_n = a_0 - a_0 r^n$$ $$S_n = \frac{a_0(1 - r^n)}{1-r}$$ Given:$$\left|r\right| &lt; 1,$$ $$\lim_{n\to \infty} S_n = \lim_{n\to \infty}\frac{a_0(1 - r^n)}{1-r} = \frac{a_0}{1-r}$$ I personally prefer Method 1 because it is faster and more intuitive, as we don't have to multiply by $r$.  Method 1 for formula of partial sums: $$\sum_{i=1}^{n}{a_0r^{i-1}} \equiv S_n$$ $$S_n = a_0r^0+a_0r^1+a_0r^2+\cdots+a_0r^{n-2}+a_0r^{n-1}$$ $$S_n = r\left(a_0r^{-1} + a_0r^{0} + a_0r^1+\cdots+a_0r^{n-3}+a_0r^{n-2}\right)$$ $$S_n = r\left(a_0r^{-1} + S_n - a_0r^{n-1}\right)$$ $$S_n = a_0 + rS_n - a_0r^{n}$$ $$(1-r)S_n = a_0 - a_0r^n$$ $$S_n = \frac{a_0(1 - r^n)}{(1-r)}$$
119958
An exercise asks me to show that if $X_1,X_2,\ldots,X_n$ is a random sample from a Poisson distribution with parameter $\theta$, the conditional distribution of $X_1,X_2,\ldots, X_{n-1}$, given $Y=\sum_{i=1}^n X_i$, is multinomial. Normally, this would have been an easy one but unfortunately the joint distribution here throws me off a bit. This is where I would appreciate some help.
107978
I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side. Please guide me how to do it further.     $1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.   Sol:  $P(n):\ 1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.  $P(1):\ \frac{1}{6}(2)(9) = \frac{1}{2}(2)(3)$.  $P(1): 3$.  Hence it is true for $n=n_0 = 1$.  Let it be true for $n=k$.  $P(k):\ 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) = \frac{1}{6}k(k+1)(2k+7)$.  We have to prove  $P(k+1):\ 1\cdot 3 + 2\cdot 4 + \cdots + (k+1)(k+3)= \frac{1}{6}(k+1)(k+2)(2k+9)$.  Taking LHS: $$\begin{align*} 1\cdot 3 + 2\cdot 4+\cdots + (k+1)(k+2) &amp;= 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) + (k+1)(k+3)\\ &amp;= \frac{1}{6}(k+1)(k+2)(2k+9) + (k+1)(k+3)\\ &amp;= \frac{1}{6}(k+1)\left[(k+2)(2k+9) + 6k+18\right]\\ &amp;= \frac{1}{6}(k+1)\left[2k^2 + 13k + 18 + 6k + 18\right]\\ &amp;= \frac{1}{6}(k+1)\left[2k^2 + 19k + 36\right]. \end{align*}$$
29310
Find the sum to $n$ terms as well as the sum to infinity of the series:  $$\frac{1}{3} + \frac{1}{4\cdot 2!} + \frac{1}{5\cdot 3!} +\cdots.$$  I was trying this question many times but I didn't get the answer. I was using geometric series formula, but it was not applicable here...I don't know where I have to start, and I don't know any hint about this series....  If anybody helps me, I would be very thankful to him.