A function is defined as [tex][tex]$a_n = a_{n-1} + \log_{n+1}(n-1)$[/tex][/tex], where [tex][tex]$a_1 = 8$[/tex][/tex]. What is the value of [tex][tex]$a_3$[/tex][/tex]?

A. 8
B. 8.5
C. 9.2
D. 10



Answer :

Let's solve for [tex]\(a_3\)[/tex] step by step using the given recurrence relation [tex]\( a_n = a_{n-1} + \log_{n+1}(n-1) \)[/tex] and the initial condition [tex]\( a_1 = 8 \)[/tex].

### Step 1: Calculate [tex]\( a_2 \)[/tex]
Given:
[tex]\[ a_1 = 8 \][/tex]

The recurrence relation for [tex]\( a_2 \)[/tex] is:
[tex]\[ a_2 = a_1 + \log_3(1) \][/tex]

Here, we need to evaluate the logarithm:
[tex]\[ \log_3(1) = 0 \][/tex]
since any number to the power of 0 is 1.

Thus:
[tex]\[ a_2 = 8 + 0 = 8 \][/tex]

### Step 2: Calculate [tex]\( a_3 \)[/tex]
Now, use the value of [tex]\( a_2 \)[/tex] to find [tex]\( a_3 \)[/tex]:
[tex]\[ a_3 = a_2 + \log_4(2) \][/tex]

Next, we need to evaluate this logarithm:
[tex]\[ \log_4(2) = \frac{\log_2(2)}{\log_2(4)} \][/tex]
[tex]\[ \log_2(2) = 1 \][/tex]
[tex]\[ \log_2(4) = 2 \][/tex]
So:
[tex]\[ \log_4(2) = \frac{1}{2} \][/tex]

Thus:
[tex]\[ a_3 = 8 + \frac{1}{2} = 8.5 \][/tex]

Therefore, the value of [tex]\( a_3 \)[/tex] is:
[tex]\[ \boxed{8.5} \][/tex]