I need nth term for the below sequence  $1,2,(3/2),(5/3),(8/5),(13/8),(21/13),(34/21)........nth term$   I know [(present number)=  sum of previous(numerator + denominator ) / previous numerator]  Thanks before
122540
$$\frac23,\frac35,\frac58,\frac8{13},\frac{13}{21},\frac{21}{34},\frac{34}{55},\ldots$$  You would be able to find here a pattern that the denominator of previous term gets new term numerator and new term denominator gets the sum of numerator and denominator of previous term
106402
The sequence of triangular numbers is well known. Consider now the triangular numbers modulo some positive natural number $k$. That is, let $T_{n,k}$ be the remainder of dividing $T_n$ by $k$. For the regular sequence of triangular numbers, we can take as convention $T_n = T_{n, \infty}$.  Question: For all values of $k, i &lt; k$, does there exists some $n$ such that $T_{n, k} = i$
61698
This is a nice problem on series convergence that I recently stumbled upon. Given a non-negative sequence of real numbers $(a_n)$ such that $$\sum_{n=1}^\infty a_n &lt; \infty,$$ show that there exists a non-decreasing sequence of non-negative numbers $b_n$ such that $$b_n \to \infty \quad\text{ and } \quad \sum_{n=1}^\infty a_n b_n &lt; \infty.$$ In other words, for every convergent series with non-negative terms, there is another convergent series with "substantially larger" terms.  I have a solution (see below), but maybe someone else has a different, simpler, and/or more elegant solution.
123656
We are given ($a_1,a_2,....,a_m$) and ($b_1, b_2,....,b_n$) sequences with non-negative integers.   Decide whether it's possible and if it is construct a matrix $\Re^{m x n}$ of ones("1") and zeros("0") where the number of ones("1") in row $i$ is $\forall i $ : $a_i$ and in column $j$ is $\forall j$ : $b_j$   Any hint is greatly appreciated.
90661
friends.  The question is:   Find the value of $ \sum_{n=1}^{\infty} \frac{n^3}{3^n} $. I know this sum converges and that it's value is $ \frac{33}{8} $, however, I can't seem to find it.  I've tried doing $3S-S$ to try and find a pattern, tried using different subtractions, all to no avail.  Any help would be gladly accepted.  Thanks in advance, Pedro
45103
I'm trying to find a recursive formula for all the ternary (using ${0,1,2}$) sequences of length $n$ which contain at least one zero, and have an even sum of digits.  My attempt so far is added below. I think the idea is right, but the recurrence relation I get fails when setting in numbers (and it's not always an integer..). any ideas?     Denote with $f\left(n\right)$ the number of legal sequences as in the question, $\bar{f}\left(n\right)$ as the number of sequences with at least one 0 with an odd number of digits, and as $g\left(n\right)$ the number of sequences with an even sum of digits and with no zeroes.      For any sequence, the sum of digits would be even if and only if there is an even number of 1s, therefore any legal sequence of length $n$ can be achieved in a unique way out of the following:      • Starting from a sequence where the digits are even and there is no zero, append zero at the end. There are $g\left(n-1\right)$ such sequences.      • Starting from a legal sequence of length $n-1$, append 0 or 2 at the end, adding $2f\left(n-1\right)$ sequences.      • Starting from a sequence with a zero where the sum of digits is odd, append 1 at the end, adding $\bar{f}\left(n-1\right)$ sequences.      These gives us that $$f\left(n\right)=2f\left(n-1\right)+\bar{f}\left(n-1\right)+g\left(n-1\right)$$      Considering $g\left(n\right)$ first, we are interested in strings of length n from $\left\{ 1,2\right\} $ with an even number of 1s. There's a simple bijection to the set of strings of length $n$ with an odd number of 1s by “flipping” the first value, and as these sets include all possibilities with no overlap, it follows that $g\left(n\right)=2^{n-1}$.      Next looking at $\bar{f}\left(n-1\right)$. We can use the fact that the sequences with a zero where the sum of digits is odd and the sequences with a zero where the sum of digits is even are exactly a disjoint union of all the sequences with a zero.      As there are $n\cdot3^{n-1}$ such sequences (we first choose one location to be 0, and fill the rest normally), we get that $\bar{f}\left(n\right)+f\left(n\right)=n\cdot3^{n-1}$, or $\bar{f}\left(n\right)=n\cdot3^{n-1}-f\left(n\right)$      Putting this all together we get that $$f\left(n\right)=2f\left(n-1\right)+\left(n-1\right)\cdot3^{n-2}-f\left(n-1\right)+2^{n-2}  or f\left(n\right)=f\left(n-1\right)+\left(n-1\right)\cdot3^{n-2}+2^{n-2}$$
113842
Given the series $A=\sum\limits_{n=1}^{+\infty}\frac{p_n}{p_{n+1}}$ and $B=\sum\limits_{n=1}^{+\infty}\frac{p_{2n}}{p_{2n+1}}$, where $p_n$ is the sequence where the nth number are the nth prime number, is it possible to determine whether they are converging or diverging?   I think there are some ways using the fact that $\sum\limits_{p\in\mathbb{P}}\frac{1}{p}$ diverges, but I'm not sure of how I can use it.
47520
While studying the Fibonacci sequence I encountered this problem in the handout, and I can not understand how to do it.     Show that if the Fibonacci sequence has a term divisible by a natural number $m$, then there are infinitely many such terms.
122547
Let ${x_n}$ be a sequence in $R$ such that $|x_{n+1} - x_n|&lt; \frac{1}{n^2}$ for all $n \in N$.  Show that the sequence is convergent.  If it were $|x_{n+1} - x_n|= \frac{1}{n^2}$, could take help of the fact $\sum \frac{1}{n^2}$ is convergent and of triangular inequality.  But what to do here? Please help. Thanks in advance.
38461
I am trying to calculate the following series:  $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$$  and I managed to reduce it to this term  $$\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$$  And here I am stuck. I tried writing down a few partial sums but I can't see the pattern, $\frac{1}{2}-\frac{1}{2}+\frac{1}{5}+\frac{1}{4}-\frac{1}{3}+\frac{1}{8}+...$ I cant seem to find a closed formula that we can calculate for $S_n$  How should I go about solving this question
42675
I'm trying to find a formula for the following sequence:  $\{\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{3\sqrt{3\sqrt{3}}},...\}$  I thought of solving it recursively and I got this formula:  $a_{n}=\sqrt{3*a_{n-1}}$  $a_{0}=1$  Is there a better and non-recursive formula for the given sequence?
44573
Say you have +1 on first iteration, +2 on second, and so on until N, and you want to know the total.  That's easily calculate using (N * (N + 1) ) / 2.  What's that equation or technique called?
106823
Let $\{ F_n\}$ be the sequence of numbers defined by $F_1=1=F_2;\, F_{n+1}=F_n+F_{n-1}$ for $n \geq 2$. Let $f_n$ be the remainder left when $F_n$ is divided by $5$. Then $f_{2000}$ equals (A) $0$ $~~~~~~~~~~~~~~~~$ (B) $1$   $~~~~~~~~~~~~~~~~$ (C) $2$$~~~~~~~~~~~~~~~~$ (D) $3$  I found that $F(1)=1$, $F(2)=1$, $F(3)=2$, $F(4)=3$, $F(5)=5$, $F(6)=8$, $F(7)=13$, $F(8)=21$, $F(9)=34$, and $F(10)=55$. But I need a systematic pattern to find $F_n$.
55299
Let $f(n)$ denote the number of sequences $a_1, a_2, \ldots, a_n$ that can be constructed where each $a_i$ is $+1$, $-1$, or $0$.  Note that no two consecutive terms can be $+1$, and no two consecutive terms can be $-1$. Prove that $f(n)$ is the nearest integer to $\frac12(1+\sqrt2)^{n+1}$.  Can anyone help me with this problem. I really do not know what to do in this one.
103553
Let $$0,0,1,0,0,1,0,0,1,...$$ be a sequence. What is the general term of this sequence?
12710
Problem :   How to find the sum of the sequence $\frac{1}{1+1^2+1^4} +\frac{2}{1+2^2+2^4} +\frac{3}{1+3^2+3^4}+.....$  I am unable to find out how to proceed in this problem.. this is a problem of arithmetic progression... Please suggest how to proceed...Thanks..
48910
So I'm being asked to find a formula for the nth fibonacci number.    I know the answer is $$x_{n}=\frac{(1+5^{1/2})^{n} -(1-5^{1/2})^n}{\sqrt{5}2^n}$$  However I don't really know how to get there.  It seems so complicated and I need some help.