To prove that f(t+3) + f(t-1) = 27f(t), we will substitute the expression for f(t) into the given equation and simplify both sides.
Starting with the left side of the equation, we have:
f(t+3) + f(t-1) = 3^(t+3) + 3^(t-1)
Since 3^(a+b) = 3^a * 3^b, we can write this as:
= 3^t * 3^3 + 3^t * 3^(-1)
Using the property that 3^(-1) = 1/3, we can rewrite this as:
= 3^t * 3^3 + 3^t * 1/3
= 3^t * 27 + 3^t/3
Since 3^t/3 can be written as 3^(t-1), we can simplify further:
= 27 * 3^t + 3^(t-1)
Now, let's simplify the right side of the equation:
27f(t) = 27 * 3^t
Now, we can compare the right side with the left side:
27 * 3^t + 3^(t-1) = 27 * 3^t
Since 27 * 3^t is already a part of the left side, we can simplify this expression to:
27 * 3^t = 27 * 3^t
Therefore, we have shown that f(t+3) + f(t-1) = 27f(t)
If f(t)=3^t, show that f(t+3)+f(t-1)=27f(t)
1 answer