Question

Find the sum of the first n odd natural number. Hence find 1+3+5+....101.

Answer 1

sum of the first n odd natural number = 2601

Step-by-step explanation:

Given

odd natural number = 1+3+5+....101

find sum = ??

Solution

here, a = 1

and common difference d = 2

sum of n numbers (Sn)

Sn = n/2 [2a + (n-1) d]

when a = 1

Sn = n/2 [2 + (n-1) 2]

Sn = n/2 (2n)

Sn = n² -----------1

since we have

1+3+5+....101

tn = 101

tn = a + (n-1) d

101 = 1 + (n-1) 2

101 - 1 = 2n - 2

100 +2 = 2n

n = 51

As Sn = n²

S51 =  51²

S51 = 2601

Answer 2

Answer:

1+3+5+....+101 is given sequence, this is an A.P.

here,

a=1 and d= 3-1=2

so, by 1st condition,

now,

Sn= n/2 [2a+(n-1)d]....(formula)

Sn=n/2 [2×1+(n-1)2]

so,Sn=n2.....................(1)

now,

by 2nd condition,

1+3+5+.....+101

so,tn=a+(n-1)d............(formula)

Tn=101

so,101=1+(n-1)2

so,n=51

put n=51 in equation 1st

Sn=n2

Sn=(51)2

Sn=2601