Question

how to show that sum of n odd terms =n^2​ correct answers please
quick
quick
quick​

Answer 1

Answer:

This is your answer

Step-by-step explanation:

S=1+3+5+⋯+(2n−5)+(2n−3)+(2n−1)is the sum of the first n odd numbersS=1+3+5+⋯+(2n−5)+(2n−3)+(2n−1)is the sum of the first n odd numbers

We can also writeWe can also write

S=(2n−1)+(2n−3)+(2n−5)+⋯+5+3+1S=(2n−1)+(2n−3)+(2n−5)+⋯+5+3+1

Adding both results in sequenceAdding both results in sequence

2 S=[1+(2n−1]+[3+(2n−3)]+[5+(2n−5)]+⋯+[(2n−5)+5]+[(2n−3)+3]+[(2n−1)+1]2 S=[1+(2n−1]+[3+(2n−3)]+[5+(2n−5)]+⋯+[(2n−5)+5]+[(2n−3)+3]+[(2n−1)+1]

⟹2 S=(2n)+(2n)+(2n)+⋯+(2n)+(2n)+(2n)n terms⟹2 S=(2n)+(2n)+(2n)+⋯+(2n)+(2n)+(2n)n terms

⟹2 S=(2n)(n)⟹2 S=(2n)(n)

⟹2 S=2n2⟹2 S=2n2

⟹2 S=n2⟹2 S=n2

⟹1+3+5+⋯+(2n−5)+(2n−3)+(2n−1)=n2