Answer
Base Case:
For n = 0, the left-hand side of the equation becomes 3, and the right-hand side becomes 3(5^1 - 1)/4 = 3. Therefore, the statement holds for n = 0.
Induction Hypothesis:
Assume that the statement is true for some positive integer k, i.e.,
3 + 3 * 5 + 3 * 5^2 + ... + 3 * 5^k = 3(5^(k+1) - 1)/4
Induction Step:
We need to prove that the statement is also true for n = k + 1, i.e.,
3 + 3 * 5 + 3 * 5^2 + ... + 3 * 5^(k+1) = 3(5^(k+2) - 1)/4
Starting with the left-hand side of the equation for n = k + 1:
3 + 3 * 5 + 3 * 5^2 + ... + 3 * 5^k + 3 * 5^(k+1)
Using the induction hypothesis:
= 3(5^(k+1) - 1)/4 + 3 * 5^(k+1)
Factoring out 3 * 5^(k+1):
= 3 * 5^(k+1) * [ (5^k - 1)/4 + 1 ]
Simplifying the expression inside the bracket:
= 3 * 5^(k+1) * [(5^k - 1 + 4)/4]
= 3 * 5^(k+1) * (5^(k+1) + 3)/4
= 3(5^(k+2) - 1)/4
Therefore, the statement holds true for n = k + 1.
Work Step by Step
Base Case:
For n = 0, the left-hand side of the equation becomes 3, and the right-hand side becomes 3(5^1 - 1)/4 = 3. Therefore, the statement holds for n = 0.
Induction Hypothesis:
Assume that the statement is true for some positive integer k, i.e.,
3 + 3 * 5 + 3 * 5^2 + ... + 3 * 5^k = 3(5^(k+1) - 1)/4
Induction Step:
We need to prove that the statement is also true for n = k + 1, i.e.,
3 + 3 * 5 + 3 * 5^2 + ... + 3 * 5^(k+1) = 3(5^(k+2) - 1)/4
Starting with the left-hand side of the equation for n = k + 1:
3 + 3 * 5 + 3 * 5^2 + ... + 3 * 5^k + 3 * 5^(k+1)
Using the induction hypothesis:
= 3(5^(k+1) - 1)/4 + 3 * 5^(k+1)
Factoring out 3 * 5^(k+1):
= 3 * 5^(k+1) * [ (5^k - 1)/4 + 1 ]
Simplifying the expression inside the bracket:
= 3 * 5^(k+1) * [(5^k - 1 + 4)/4]
= 3 * 5^(k+1) * (5^(k+1) + 3)/4
= 3(5^(k+2) - 1)/4
Therefore, the statement holds true for n = k + 1.
