SOLUTION: Compute \sum_{k = 1}^{100} k (\lceil \log_3 k \rceil

SOLUTION: Compute \sum_{k = 1}^{100} k (\lceil \log_3 k \rceil - \lfloor \log_9 k \rfloor).

Question 1209893: Compute
\sum_{k = 1}^{100} k (\lceil \log_3 k \rceil - \lfloor \log_9 k \rfloor).
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this summation step-by-step.
**Understanding the Terms**
We have the summation:
$$\sum_{k=1}^{100} k (\lceil \log_3 k \rceil - \lfloor \log_9 k \rfloor)$$
* **$\lceil \log_3 k \rceil$:** This represents the smallest integer greater than or equal to $\log_3 k$.
* **$\lfloor \log_9 k \rfloor$:** This represents the largest integer less than or equal to $\log_9 k$.
**Key Property:**
We know that $\log_9 k = \log_{3^2} k = \frac{1}{2} \log_3 k$.
**Analyzing the Difference**
Let's look at the difference $\lceil \log_3 k \rceil - \lfloor \log_9 k \rfloor$.
1. **If $\log_3 k$ is an integer:**
* Then $\lceil \log_3 k \rceil = \log_3 k$.
* $\lfloor \log_9 k \rfloor = \lfloor \frac{1}{2} \log_3 k \rfloor$.
* The difference will be $\log_3 k - \lfloor \frac{1}{2} \log_3 k \rfloor$.
2. **If $\log_3 k$ is not an integer:**
* We still have $\lfloor \log_9 k \rfloor = \lfloor \frac{1}{2} \log_3 k \rfloor$.
**Simplifying the Difference**
Let $x = \log_3 k$. Then we want to understand $\lceil x \rceil - \lfloor x/2 \rfloor$.
* If $x$ is an integer, then $\lceil x \rceil = x$.
* If $x$ is not an integer, then $\lceil x \rceil = \lfloor x \rfloor + 1$.
Let's consider possible values of $\lfloor x/2 \rfloor$.
* If $x = 2n$, where $n$ is an integer, then $\lfloor x/2 \rfloor = n$, and $\lceil x \rceil = 2n$. The difference is $2n - n = n = x/2$.
* If $x = 2n + 1$, where $n$ is an integer, then $\lfloor x/2 \rfloor = n$, and $\lceil x \rceil = 2n + 1$. The difference is $2n + 1 - n = n + 1 = (x+1)/2$.
* If $2n < x < 2n+1$, then $\lfloor x/2 \rfloor = n$. The difference is $\lceil x \rceil - n$.
* If $2n+1 < x < 2n+2$, then $\lfloor x/2 \rfloor = n$. The difference is $\lceil x \rceil - n$.
The difference $\lceil \log_3 k \rceil - \lfloor \log_9 k \rfloor$ is 1 if $\log_3 k$ is not an even integer, and is 0 if $\log_3 k$ is an even integer.
**When is the difference 1?**
The difference is 1 when $\log_3 k$ is not an even integer. This occurs when $k$ is not a power of $3^2 = 9$.
**When is the difference 0?**
The difference is 0 when $\log_3 k$ is an even integer. This occurs when $k$ is a power of $9$.
**Calculating the Sum**
We need to calculate $\sum_{k=1}^{100} k (\lceil \log_3 k \rceil - \lfloor \log_9 k \rfloor)$. This is equivalent to summing all $k$ values except for powers of $9$.
* Powers of 9 in the range: $1, 9, 81$.
* Sum of all integers from 1 to 100: $\frac{100(101)}{2} = 5050$.
* Sum of powers of 9: $1 + 9 + 81 = 91$.
The desired sum is $5050 - 91 = 4959$.
**Final Answer:** The final answer is $\boxed{4959}$

RELATED QUESTIONS
log(3)k + log(3)k-2 =... (answered by ewatrrr)

How to compute {{{ sum( ((-1)^k * k ), k=1, 50)... (answered by ikleyn)

let log k 2=11/9. find log 8... (answered by stanbon)

Find the integer k, k > 2, for, which log (k - 2)! + log(k - 1)! + 2 = 2 log... (answered by richard1234)

log(5)k - log(5)k-1 =... (answered by LinnW)

Please solve : log (k+2) + log (k-1) =... (answered by ewatrrr,srinivas.g)

{{{sum(k^3,k=4,24)}}} (answered by MathLover1,Edwin McCravy)

Sum of (1, 5 k(k+1)/2, 25 k(k+1)(k+2)/6, 125k(k+1)(k+2) (k+3)/24, ��... (answered by robertb)

9(k-10)=-k (answered by checkley71)