All Questions
25
questions
6
votes
1
answer
105
views
calcuate $\sum_{i=0}^{n} 2^{2i}$
I want to calcuate this problem: $\sum_{i=0}^{n} 2^{2i+5}$
I know that we can expand this problem like this:
$\sum_{i=0}^{n} (2^{2i+5})$
$=\sum_{i=0}^{n} (2^5 \times 2^{2i})$
$=\sum_{i=0}^{n} (32 \...
- 337
0
votes
2
answers
187
views
find the first term of the series?
The sum of an infinite geometric series of real numbers is $14,$ and the
sum of the cubes of the terms of this series is $392$.
What is the first term of the series?
My attempt: Let the series be $\...
- 14.6k
0
votes
5
answers
334
views
Find the sum to $n$ terms of the following series
Find the sum to $n$ terms of the following series:
$$\dfrac {2}{5}+\dfrac {6}{5^2}+\dfrac {10}{5^3}+\dfrac {14}{5^4}+………$$
My Attempt:
Let
$$S_n=\dfrac {2}{5} + \dfrac {6}{5^2}+\dfrac {10}{5^3}+\...
- 7,426
56
votes
7
answers
8k
views
Can we calculate $ i\sqrt { i\sqrt { i\sqrt { \cdots } } }$?
It might be obvious that $2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { \cdots } } } } } } $ equals $4.$ So what about $i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } } ...
user536517
3
votes
3
answers
153
views
Summing up $3+5+9+17+...$
Find the sum of sum of $3
+5+9+17+...$ till $n$ terms.
Using Method of differences, the sum of the series is
$$\sum\limits_{j=1}^n 2^{j-1}+n$$
I am facing difficulty in evaluating $$\sum\limits_{j=1}...
- 6,111
-2
votes
3
answers
131
views
Induction proof of $1+3+\cdots+3^n=\frac{3^{n+1}-1}{2}$ [closed]
How would I prove the following by induction?$$1+3+3^2+3^3+\cdots+3^n=\frac{3^{n+1}-1}{2}$$
for all $n\geq 0.$
I kept trying to create a base case but I am not sure how many I need. I also seem to be ...
- 2,042
6
votes
4
answers
577
views
Find closed formula by changing order of summation: $\sum_{i=1}^ni3^i$
Working on homework for a probability and computing class, but my ability to work with summations is rusty to say the least, so I suspect this is going to turn out pretty straightforward.
Problem ...
- 73
0
votes
1
answer
2k
views
Range of values of x for the sum to be valid.
Given the series:
$$\frac{1}{x+1}+\frac{1}{(x+1)^2}+\frac{1}{(x+1)^3}+...$$
Find the sum to infinity for the series and state the range(s) of values of $x$ for the sum to be valid.
I solved the ...
- 7,798
4
votes
1
answer
687
views
$a \in (0,1]$ satisfies $a^{2008} -2a +1 = 0$ and we define $S$ as $S=1+a+a^2+a^3........a^{2007}$. The sum of all possible value(s) of $S$ is?
This is homework.
Let $a \in (0,1]$ satisfies the equation $$a^{2008} -2a +1 = 0$$
and we define $S$ as $$S=1+a+a^2+a^3........a^{2007}$$
The sum of all possible value(s) of $S$ is?
My Attempt
$...
- 2,944
3
votes
3
answers
397
views
How to calculate sum of series (Geometric?): $\sum_{n=0}^{\infty}\frac 12 \left(\frac14\right)^{n-1}$
$$\sum_{n=0}^{\infty}\frac 12 \left(\frac14\right)^{n-1}.$$
This is what I tried to do
$$\frac 12\sum_{n=0}^\infty \left(\frac14\right)^{n-1} =\frac 12 \sum_{n=0}^\infty \left(-\frac14\right) \left(\...
- 927
0
votes
0
answers
68
views
Polynomials and power series. How to prove $\frac{1-x^{11}}{1-x}=1+x+x^2+x^3+\cdots+x^{10}$
How do I prove that $$\dfrac{1-x^{11}}{1-x}=1+x+x^2+x^3+\cdots+x^{10} $$
Attempt: By observing first let assume that $∣x∣<1$ then $\dfrac{1-x^{11}}{1-x}=\dfrac{1}{1-x}(1-x^{11})=(1+x+x^2+x^3+x^4+\...
- 2,034
2
votes
5
answers
6k
views
How do I prove that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})$
I am unsuccessfully attempting a problem from Spivak's popular book 'Calculus' 3rd edition. The problem requires proof for the following equation:
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}+\dotsb+xy^{n-2}+y^{n-...
- 23
11
votes
1
answer
13k
views
Sum of series : $1+11+111+...$
Sum of series $1+11+111+\cdots+11\cdots11$ ($n$ digits)
We have:
$1=\frac {10-1}9,$
$11=\frac {10^2-1}9$
.
.
.
$11...11= \frac {10^n-1}9$ (number with $n$ digits)
and summing them we find the ...
- 11.2k
1
vote
2
answers
11k
views
Proof for formula for sum of sequence $3 + 9 + 27 + 81...+ 3^n$
Trying to formulate a proof for that sequence as practice. After reading this question's answer and lecture on this, I decided to try and practice with this sequence.
My try:
Base case $n = 1$
$S(n) ...
- 25
1
vote
7
answers
209
views
Why is this true $\frac{1-y^n}{1-y}=(1+y+y^2+...+ y^{n-1})$? [closed]
I have a heard time seeing why is this true
$\frac{1-y^n}{1-y}=(1+y+y^2+...+ y^{n-1})$
Could you show me some kind of proof, or an identity that would me to find this?
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