Sequence and series Class 12 Math Notes | Exercise - 4.2 (NEW CURRICULUM)

Complex Number Class 12 Mathematics Notes has been updated according to the latest syllabus of 2080. It means the solutions of Complex Number chapter provided in this channel contains all the new exercise that has recently been updated. Now you don’t need to go anywhere searching for the notes of Complex Number because we are here to serve you.
*What is Mathematical Induction?*
Mathematical induction is a method used to prove that a statement is true for all positive integers. It's like climbing a staircase, where you start at the first step and prove that the statement is true for that step. Then, you assume that the statement is true for some arbitrary step, and prove that it's also true for the next step. By doing so, you can conclude that the statement is true for all positive integers.
*Example:*
Let's say we want to prove that 2^n is grater than n for all positive integers n.
1. *Base Case:* We can prove that 2^1 is grater than1, which is true.
2. *Inductive Step:* Assume that 2^k is grater than k is true for some arbitrary positive integer k.
3. *Proof:* We need to show that 2^(k+1) is grater than (k+1) is also true. We can do this by writing:
2^(k+1) = 2(2^k) is grater than 2k (since we assumed 2^k is grater than. k )
So, we have:
2^(k+1) is grater than 2k is grater tan k+1 (since k+1is grater than equal 2k)
This shows that 2^(k+1) is grater than (k+1), which completes the inductive step.
*Conclusion:*
Since we've proven the base case and the inductive step, we can conclude that 2^n is grater than n is true for all positive integers n.
*Tips and Tricks:*
* Always start with the base case to ensure that your proof is sound.
* In the inductive step, assume that the statement is true for some arbitrary positive integer k.
* Try to simplify your proof by using algebraic manipulations or visual aids.
* Practice makes perfect! Try solving more problems using mathematical induction.
*The Mathematical Induction Principle:*
Let P(n) be a statement about a positive integer n. If:
1. P(1) is true (the base case)
2. P(k) is true implies P(k+1) is true (the inductive step)
Then, P(n) is true for all positive integers n.
*Formula:*
P(n) ∴ P(n+1) (for all positive integers n)
*Example:*
Prove that 1 + 2 + 3 + ... + n = n(n+1)/2 for all positive integers n.
*Base Case:* For n = 1, we have 1 = 1(1+1)/2, which is true.
*Inductive Step:* Assume that the statement is true for some arbitrary positive integer k. We need to show that it is also true for k+1.
Let's write out the sum:
1 + 2 + 3 + ... + k + (k+1)
Using the inductive hypothesis, we can rewrite this as:
k(k+1)/2 + (k+1)
Now, we can simplify and rewrite this as:
(k+1)(k/2+1)
This shows that the statement is true for k+1, completing the inductive step.
*Conclusion:* Since we've proven the base case and the inductive step, we can conclude that the statement 1 + 2 + 3 + ... + n = n(n+1)/2 is true for all positive integers n.
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class 12 math sequence and series exercise 4.2 notes new curriculum.