se Mathematical Induction to prove that for all n ≥ 1,
se Mathematical Induction to prove that for all n ≥ 1,
1 +
1
3
+
1
3
2
+
1
3
3
+ · · · +
1
3
2n
=
3
2
�
1 − (
1
3
)
2n+1�
2. Let f(x) = x
3+9x
x2−3x−4
To prove the given statement using mathematical induction, we first establish the base case for n=1:
1 + (1/3) = 3/2(1 - (1/3)^2(1)) = 3/2(1 - 1/9) = 3/2(8/9) = 1
So the base case holds true.
Next, we assume that the statement is true for some arbitrary positive integer k, i.e.,
1 + (1/3) + (1/3)^2 + ... + (1/3)^(2k) = 3/2(1 - (1/3)^(2k+1))
Now we aim to prove it for k + 1:
1 + (1/3) + (1/3)^2 + ... + (1/3)^(2k) + (1/3)^(2k+1) = 3/2(1 - (1/3)^(2k+2))
Adding (1/3)^(2k+1) to both sides of the assumption, we get:
1 + (1/3) + (1/3)^2 + ... + (1/3)^(2k) + (1/3)^(2k+1) = 3/2(1 - (1/3)^(2k+1)) + (1/3)^(2k+1)
= 3/2(1 - (1/3)^(2k+1)) + 1/3 *(1/3)^(2k+1)
= 3/2(1 - (1/3)^(2k+1)) + (1/3)^(2k+2)
= 3/2(1 - (1/3)^(2k+2))
This completes the induction step, and we have proven the statement using mathematical induction.