If `t_(n)=(1+(-2)^(n))/(n-1)`, find `t_(6)-t_(5)`

Answer

To solve the problem, we need to find \( t_6 - t_5 \) using the formula given for \( t_n \):

\[ t_n = \frac{1 + (-2)^n}{n - 1} \]

### Step 1: Calculate \( t_6 \)

Substituting \( n = 6 \) into the formula:

\[ t_6 = \frac{1 + (-2)^6}{6 - 1} \]

Calculating \( (-2)^6 \):

\[ (-2)^6 = 64 \]

Now substituting back into the equation:

\[ t_6 = \frac{1 + 64}{5} = \frac{65}{5} = 13 \]

### Step 2: Calculate \( t_5 \)

Now substituting \( n = 5 \) into the formula:

\[ t_5 = \frac{1 + (-2)^5}{5 - 1} \]

Calculating \( (-2)^5 \):

\[ (-2)^5 = -32 \]

Now substituting back into the equation:

\[ t_5 = \frac{1 - 32}{4} = \frac{-31}{4} \]

### Step 3: Calculate \( t_6 - t_5 \)

Now we can find \( t_6 - t_5 \):

\[ t_6 - t_5 = 13 - \left(-\frac{31}{4}\right) \]

This simplifies to:

\[ t_6 - t_5 = 13 + \frac{31}{4} \]

To combine these, we convert 13 into a fraction with a denominator of 4:

\[ 13 = \frac{52}{4} \]

Now we can add the fractions:

\[ t_6 - t_5 = \frac{52}{4} + \frac{31}{4} = \frac{52 + 31}{4} = \frac{83}{4} \]

### Final Answer

Thus, the final answer is:

\[ t_6 - t_5 = \frac{83}{4} \]

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