$$ \sum _{k=1}^{4}(3x+2k)+\sum _{t=2}^{12}2(3)^{t-2}=\sum _{p=1}^{\infty }(\frac{5}{2}(\frac{1}{2}))^{p} $$

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We start with the equation

$$
\sum_{k=1}^{4}(3x+2k)+\sum_{t=2}^{12}2\cdot 3^{\,t-2}=\sum_{p=1}^{\infty}\left(\frac{5}{2}\cdot\frac{1}{2}\right)^{p}.
$$

**Step 1. Evaluate the first sum on the left-hand side**

The sum is

$$
\sum_{k=1}^{4}(3x+2k).
$$

We can split the sum into two parts:

$$
\sum_{k=1}^{4} 3x + \sum_{k=1}^{4} 2k.
$$

Since $$3x$$ does not depend on $$k$$, we have

$$
\sum_{k=1}^{4} 3x = 4\cdot 3x = 12x.
$$

Next, we evaluate

$$
\sum_{k=1}^{4}2k = 2\sum_{k=1}^{4}k.
$$

The sum $$\sum_{k=1}^{4}k$$ is

$$
1+2+3+4 = 10.
$$

Therefore,

$$
2\sum_{k=1}^{4}k = 2\cdot 10 = 20.
$$

Thus, the first sum becomes

$$
12x + 20.
$$

---

**Step 2. Evaluate the second sum on the left-hand side**

The sum is

$$
\sum_{t=2}^{12}2\cdot 3^{t-2}.
$$

Introduce a change of variable by letting

$$
j = t-2.
$$

When $$t=2$$, $$j=0$$, and when $$t=12$$, $$j=10$$. Thus the sum becomes

$$
2\sum_{j=0}^{10}3^{\,j}.
$$

The sum of a geometric series $$\sum_{j=0}^{10}3^{\,j}$$ is given by

$$
\sum_{j=0}^{10}3^{\,j} = \frac{3^{11}-1}{3-1} = \frac{3^{11}-1}{2}.
$$

Multiplying by $$2$$, we have

$$
2\cdot \frac{3^{11}-1}{2} = 3^{11}-1.
$$

Thus the second sum is

$$
3^{11}-1.
$$

---

**Step 3. Combine the left-hand side sums**

Adding the two parts, the left-hand side (LHS) of the equation is

$$
12x + 20 + (3^{11}-1) = 12x + 3^{11} + 19.
$$

For reference, $$3^{11}$$ is a large number, specifically

$$
3^{11}=177147.
$$

So the LHS can also be written as

$$
12x + 177147 + 19 = 12x + 177166.
$$

---

**Step 4. Evaluate the right-hand side**

The right-hand side (RHS) is

$$
\sum_{p=1}^{\infty}\left(\frac{5}{2}\cdot\frac{1}{2}\right)^{p}.
$$

Simplify the expression inside the sum:

$$
\frac{5}{2}\cdot\frac{1}{2}=\frac{5}{4}.
$$

So the series becomes

$$
\sum_{p=1}^{\infty}\left(\frac{5}{4}\right)^{p}.
$$

This is an infinite geometric series with first term

$$
a = \frac{5}{4}
$$

and common ratio

$$
r = \frac{5}{4}.
$$

For an infinite geometric series to converge, we must have $$|r|<1$$. Here,

$$
\left|\frac{5}{4}\right|=\frac{5}{4}>1.
$$

Since the common ratio exceeds 1, the series diverges, meaning its sum is infinite.

---

**Step 5. Analyze the equation**

The equation now becomes

$$
12x+177166=\infty.
$$

For any finite value of $$x$$, the left-hand side is a finite number, whereas the right-hand side is infinite. Therefore, there is no finite value of $$x$$ that can satisfy the equation.

---

**Final Conclusion**

No solution exists for $$x$$ since the right-hand side diverges.
There is no solution for $$ x $$ because the right-hand side of the equation is infinite, making the equation unsolvable for any finite $$ x $$.

\( \sum _{k=1}^{4}(3x+2k)+\sum _{t=2}^{12}2(3)^{t-2}=\sum _{p=1}^{\infty }(\frac{5}{2}(\frac{1}{2}))^{p} \)

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