00:01
Using the mathematical induction method, we have to show that for all n greater than or equal to 1, summation j is running from 1 to n, 4j plus 1, it will be equal to 2n square plus 3n.
00:18
We have to show this result.
00:20
So for the proof of this result, we will take the steps.
00:25
So step first is nothing but the base case, that is when n is equal to 1.
00:31
So consider lhs of the thing which we have to prove.
00:35
We have to prove this star.
00:36
So for this lhs, it is equal to summation j is running from 1 to n, 4j plus 1.
00:44
For n it is equal to 1, it is equal to 4 into 1 plus 1.
00:49
So here n is equal to 1.
00:52
So 4 plus 1, it is equal to 5.
00:56
Now consider the rhs.
00:58
Rhs is nothing but twice n square plus thrice n, where n it is equal to 1, that is 2 plus 3 which is equal to 5.
01:06
That is rhs is nothing but equal to the lhs.
01:10
Therefore, case 1, that is n equal to 1, base case holds here.
01:19
Now step 2 is nothing but the inductive hypothesis.
01:22
Now let's assume that this statement is true for some positive integer k.
01:28
That means summation j is running from 1 to k, 4j plus 1.
01:35
It is equal to twice k square plus thrice k.
01:40
By induction hypothesis, this is true.
01:43
Now in the step 3, we will take the step 3 as a inductive step.
01:49
We have to show that this statement is true for n it is equal to k plus 1.
01:55
We will prove it as follows that summation j is running from 1 to k plus 1 and its value is 4j plus 1.
02:05
We are just writing k plus 1, that is summation j is equal to 1 to k.
02:13
We are just writing k plus 1 here.
02:15
So summation j equal to 1 to k of 4j plus 1 plus k plus 1th term, k plus 1 term is nothing but j is replaced by k plus 1.
02:26
So here k plus 1 and 4j plus 1, this is our 4k plus 1 plus 1 is the k plus 1th term.
02:35
This is the expansion.
02:36
Now using the induction hypothesis, we can replace the first part...