Sequency is represented $$ \sum_{p=0}^{k}\left(\frac{1}{3} p+1 / 6\right)=201 / 6 $$

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It looks like you're working with the following summation:

$$
\sum_{p=0}^{k}\left(\frac{1}{3} p + \frac{1}{6}\right) = \frac{201}{6}
$$

Let's break down the problem to find the value of $$ k $$.

## Step 1: Simplify the Summation

First, let's separate the summation into two parts:

$$
\sum_{p=0}^{k} \left(\frac{1}{3} p + \frac{1}{6}\right) = \frac{1}{3} \sum_{p=0}^{k} p + \frac{1}{6} \sum_{p=0}^{k} 1
$$

### Sum of $$ p $$ from 0 to $$ k $$:

$$
\sum_{p=0}^{k} p = \frac{k(k + 1)}{2}
$$

### Sum of 1 from 0 to $$ k $$:

$$
\sum_{p=0}^{k} 1 = k + 1
$$

### Substitute Back:

$$
\frac{1}{3} \cdot \frac{k(k + 1)}{2} + \frac{1}{6} (k + 1) = \frac{201}{6}
$$

Simplify the left side:

$$
\frac{k(k + 1)}{6} + \frac{k + 1}{6} = \frac{201}{6}
$$

Factor out $$ \frac{k + 1}{6} $$:

$$
\frac{(k + 1)(k + 1)}{6} = \frac{201}{6}
$$

$$
\frac{(k + 1)^2}{6} = \frac{201}{6}
$$

## Step 2: Solve for $$ k $$

Multiply both sides by 6 to eliminate the denominator:

$$
(k + 1)^2 = 201
$$

Take the square root of both sides:

$$
k + 1 = \sqrt{201}
$$

$$
k = \sqrt{201} - 1
$$

Calculate the value:

$$
\sqrt{201} \approx 14.177
$$

$$
k \approx 14.177 - 1 = 13.177
$$

## Conclusion

The value of $$ k $$ that satisfies the equation is approximately **13.177**.

**Note:** Since $$ k $$ typically represents an integer in summations, there is no integer value of $$ k $$ that will make the summation exactly equal to $$ \frac{201}{6} $$. The closest integer values are:

- For $$ k = 13 $$:
  $$
  \sum_{p=0}^{13}\left(\frac{1}{3} p + \frac{1}{6}\right) = \frac{(13 + 1)^2}{6} = \frac{196}{6} \approx 32.666...
  $$

- For $$ k = 14 $$:
  $$
  \sum_{p=0}^{14}\left(\frac{1}{3} p + \frac{1}{6}\right) = \frac{(14 + 1)^2}{6} = \frac{225}{6} = 37.5
  $$

Neither of these sums exactly equals $$ \frac{201}{6} = 33.5 $$, but $$ k $$ around 13.177 is the precise solution if non-integer values are acceptable.
The value of $$ k $$ is approximately **13.177**.

Sequency is represented \( \sum_{p=0}^{k}\left(\frac{1}{3} p+1 / 6\right)=201 / 6 \)

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