Find the sum of the series ` 7 + 10 + 22 + 70 + . . . `up to n terms

Answer

To find the sum of the series \( 7 + 10 + 22 + 70 + \ldots \) up to \( n \) terms, we will follow these steps:

### Step 1: Identify the terms of the series
The given series is:
- First term \( T_1 = 7 \)
- Second term \( T_2 = 10 \)
- Third term \( T_3 = 22 \)
- Fourth term \( T_4 = 70 \)

### Step 2: Find the differences between consecutive terms
We calculate the differences:
- \( T_2 - T_1 = 10 - 7 = 3 \)
- \( T_3 - T_2 = 22 - 10 = 12 \)
- \( T_4 - T_3 = 70 - 22 = 48 \)

So, the differences are \( 3, 12, 48 \).

### Step 3: Find the ratios of the differences
Now, we find the ratios of these differences:
- \( \frac{12}{3} = 4 \)
- \( \frac{48}{12} = 4 \)

Since the ratios are constant, the differences of the terms are in a geometric progression (GP).

### Step 4: Write the general term of the series
Since the differences are in GP, we can express the \( n \)-th term of the series as:
\[ T_n = a \cdot r^{n-1} + b \]
where \( a \) is a constant, \( r \) is the common ratio, and \( b \) is another constant.

### Step 5: Determine the constants \( a \) and \( b \)
Using the first term \( T_1 = 7 \):
\[ T_1 = a \cdot r^{1-1} + b = a + b = 7 \quad \text{(Equation 1)} \]

Using the second term \( T_2 = 10 \):
\[ T_2 = a \cdot r^{2-1} + b = a \cdot r + b = 10 \quad \text{(Equation 2)} \]

Using the third term \( T_3 = 22 \):
\[ T_3 = a \cdot r^{3-1} + b = a \cdot r^2 + b = 22 \quad \text{(Equation 3)} \]

### Step 6: Solve the equations
From Equation 1:
\[ b = 7 - a \]

Substituting \( b \) in Equation 2:
\[ a \cdot r + (7 - a) = 10 \implies ar - a = 3 \implies a(r - 1) = 3 \quad \text{(Equation 4)} \]

Substituting \( b \) in Equation 3:
\[ a \cdot r^2 + (7 - a) = 22 \implies ar^2 - a = 15 \implies a(r^2 - 1) = 15 \quad \text{(Equation 5)} \]

### Step 7: Divide Equation 5 by Equation 4
\[ \frac{a(r^2 - 1)}{a(r - 1)} = \frac{15}{3} \implies \frac{r^2 - 1}{r - 1} = 5 \]
Factoring the left side:
\[ (r + 1) = 5 \implies r = 4 \]

### Step 8: Substitute \( r \) back to find \( a \) and \( b \)
Using \( r = 4 \) in Equation 4:
\[ a(4 - 1) = 3 \implies 3a = 3 \implies a = 1 \]
Now substituting \( a \) back to find \( b \):
\[ b = 7 - 1 = 6 \]

### Step 9: Write the general term
Thus, the \( n \)-th term is:
\[ T_n = 1 \cdot 4^{n-1} + 6 = 4^{n-1} + 6 \]

### Step 10: Find the sum of the first \( n \) terms
The sum \( S_n \) of the first \( n \) terms is given by:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (4^{k-1} + 6) \]
This can be separated into two sums:
\[ S_n = \sum_{k=1}^{n} 4^{k-1} + \sum_{k=1}^{n} 6 \]
The first sum is a geometric series:
\[ \sum_{k=1}^{n} 4^{k-1} = \frac{1(4^n - 1)}{4 - 1} = \frac{4^n - 1}{3} \]
The second sum is:
\[ \sum_{k=1}^{n} 6 = 6n \]
Combining these:
\[ S_n = \frac{4^n - 1}{3} + 6n \]

### Final Result
Thus, the sum of the series up to \( n \) terms is:
\[ S_n = \frac{4^n - 1}{3} + 6n \]