Question

Sum of n odd number through ap

Answer 1

Odd no.s form am A.P. The sum of A.P is given by the formula :-

Sn = n/2 (2a + (n-1)d) where a= first term , n= no. of terms , d= difference

In case of odd no.s 1 + 3 + 5+……..+n

a =3 , no. of terms =n and d=2

Putting the values in above formula we get,

Sn = n/2 (2×3 + (n-1)2)

=n/2(6 + 2n-2)

=n/2( 2n+ 4)

=n(n+2)

Or n² + 2n

Therefore the sum is n²+ 2n.

12.9k Views · View 8 Upvoters

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OTHER ANSWERS



LAXMINARAYAN SAHU, works at Thermal Power Plants

Answered Mar 10, 2016

Sum of n odd numbers:

suppose we take following n numbers-

=1+3+5+7

=1+(2+1)+(3+2)+(4+3)          we can write again as..

=[1+2+3+4] +[1+2+3]

=[Sum of n natural numbers]+[Sum of (n-1) natural numbers]

=n(n+1)/2 + (n-1) [(n-1)+1]/2    By solving it,we find

=[n^2+n]/2+[n^2-n]/2      again we can simplify as

=(2n^2)/2

=n^2

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Palaash Barot, works at Schools (2005-present)

Answered Apr 20, 2017 · Author has 80answers and 128.7k answer views

For Amy A.P(Arithmetic Progression), the sum of numbers is given by,

Sn=1/2×n[2a+(n-1)×d]

Where,

Sn= Sum of n numbers

n = n numbers

a = First term of an A.P

d= Common difference in an A.P

But, to find the sum of n odd numbers, there is a quick formulae,

Sn(odd numbers)= n²

For instance, what is the sum of first 10 odd numbers??

Here, n=10

Therefore, Sn=n²

Sn = 10²

Sn=100

Hope that helps!!