Final answer:
To prove the given formula is true for all natural numbers n, use mathematical induction. Verify the base case, assume the formula holds true for k, prove it holds true for k + 1, and conclude by mathematical induction.
Step-by-step explanation:
To prove the given formula is true for all natural numbers n, we will use mathematical induction.
Step 1:
Base Case: Verify the formula for n = 1.
Substituting n = 1 into the formula gives us:
1.2 = 2[1 + (1-1)2]
2 = 2[1 + 0]
2 = 2
The formula holds true for n = 1.
Step 2:
Inductive Hypothesis: Assume the formula holds true for some positive integer k, i.e.,
1.2+2.2²+3.2³+...+k.2^k = 2[1+(k-1)2^k]
Step 3:
Inductive Step: Prove the formula holds true for k + 1.
We need to prove that:
1.2+2.2²+3.2³+...+k.2^k+(k+1).2^(k+1) = 2[1+k.2^k]
Using the inductive hypothesis, we can rewrite the left side as:
2[1+(k-1)2^k] + (k+1).2^(k+1)
Expanding and simplifying the expression:
2 + 2(k-1)2^k + (k+1)2^(k+1)
= 2 + 2k.2^k - 2^k + (k+1)2^(k+1)
Factoring out 2^k:
= 2[1 + k.2^k] - 2^k + (k+1)2^(k+1)
Using the formula for the sum of a geometric series, we have:
= 2[1 + k.2^k] - 2^k + 2(k+1)2^k
= 2[1 + k.2^k - 2^k + (k+1)2^k]
= 2[1 + (k+1).2^k]
The formula holds true for k + 1.
Step 4:
Conclusion: By mathematical induction, the formula is true for all natural numbers n.