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If x >– 1, x ≠ 0, then the statement P(n): (1 + x)n > 1 + nx, n ∈ N is true for
- a)all n ∈ N
- b)all n ∈ N − {1}
- c)all n ∈ N − {2}
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Ifx >– 1,x ≠ 0, then the statementP(n): (1 + x)n > 1 + ...
P(1): (1 + x)1 > 1 + x is not true.
For P(2): (1 + x)2 > 1 + 2x is true as x ≠ 0
Let P(k): (1 + x)k > 1 + kx be true
⇒ (1 + x)(1 + x)k > (1 + x) (1 + kx)
⇒ (1 + x) k + 1 > 1 + (k + 1)x + kx2
⇒ (1 + x)k + 1 > 1 + (k + 1)x (∵ kx2 > 0)
⇒ P(k + 1) is true.
Hence (B) is the correct answer.
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Community Answer
Ifx >– 1,x ≠ 0, then the statementP(n): (1 + x)n > 1 + ...
Explanation:
Given Statement:
If x > – 1, x ≠ 0, then the statement P(n): (1 + x)n > 1 + nx, n ∈ N is true.
Proof:
1. Base Case:
Let's check the base case for n = 1:
(1 + x)1 = 1 + x and 1 + 1x = 1 + x
Since x ≠ 0, the inequality holds for n = 1.
2. Inductive Step:
Assume the statement P(k) is true for some positive integer k ∈ N (Inductive Hypothesis):
(1 + x)k > 1 + kx
Now, we need to prove that P(k + 1) is also true:
(1 + x)k+1 = (1 + x)(1 + x)k = (1 + x) + x(1 + x)k
Since x > – 1, x ≠ 0, and the inductive hypothesis (1 + x)k > 1 + kx, we have:
(1 + x) + x(1 + x)k > 1 + kx + x(1 + x)k
Expanding and simplifying:
(1 + x) + x(1 + x)k > 1 + kx + x + kx^2
1 + x + x + x^2k > 1 + kx + x + kx^2
1 + x(1 + x)k > 1 + (k + 1)x
Therefore, P(k + 1) is true if P(k) is true.
Conclusion:
By the principle of mathematical induction, the statement P(n): (1 + x)n > 1 + nx is true for all positive integers n ∈ N − {1}. Hence, the correct answer is option 'B'.
Given Statement:
If x > – 1, x ≠ 0, then the statement P(n): (1 + x)n > 1 + nx, n ∈ N is true.
Proof:
1. Base Case:
Let's check the base case for n = 1:
(1 + x)1 = 1 + x and 1 + 1x = 1 + x
Since x ≠ 0, the inequality holds for n = 1.
2. Inductive Step:
Assume the statement P(k) is true for some positive integer k ∈ N (Inductive Hypothesis):
(1 + x)k > 1 + kx
Now, we need to prove that P(k + 1) is also true:
(1 + x)k+1 = (1 + x)(1 + x)k = (1 + x) + x(1 + x)k
Since x > – 1, x ≠ 0, and the inductive hypothesis (1 + x)k > 1 + kx, we have:
(1 + x) + x(1 + x)k > 1 + kx + x(1 + x)k
Expanding and simplifying:
(1 + x) + x(1 + x)k > 1 + kx + x + kx^2
1 + x + x + x^2k > 1 + kx + x + kx^2
1 + x(1 + x)k > 1 + (k + 1)x
Therefore, P(k + 1) is true if P(k) is true.
Conclusion:
By the principle of mathematical induction, the statement P(n): (1 + x)n > 1 + nx is true for all positive integers n ∈ N − {1}. Hence, the correct answer is option 'B'.
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