Question

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What is the sum of first 'n' odd positive integers ?


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Answer 1

We know that the nth odd number is 2n−1 and the nth even number is 2n.

Now, let's assume the sum of first n odd numbers to be S i.e. S=1+3+5+…+(2n−1)

Now let us add 1 n times to the right side,

S+n=(1+1)+(3+1)+(5+1)+…+(2n−1+1)

or,S+n=2+4+6+…+2n

Now adding these 2 equations, we get,

2S+n=1+2+3+…+(2n−1)+2n

The RHS is the sum of first 2n natural numbers which is as below,

2S+n=2n(2n+1)/2

or,2S+n=2n2+n

or,2S=2n2

or,S=n2

So the sum of first n odd integers is n2.

Answer 2

The sum of first odd positive number is n^2

the series will be
1+3+5+7+9+………+n

here n= 2n-1

since the last term will be 2n-1
difference=1
and first term =1

sum = n(1+ 2n-1) /2
=2n^2/2
= n^2

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