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2
Solution: We know that the sum of first n odd natural numbers is n .
2
(a) 1 + 3 + 5 + 7 + 9 = 5 = 25
2
(b) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10) = 100
2
(c) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = (12) = 144
Example 14: Write (a) 49 and (b) 169 as a sum of odd natural numbers.
Solution: (a) As we know that 49 = 7 2

So, it can be written as:
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

(b) As we know that 169 = (13) 2
So, it can be written as:

169 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25
Example 15: Express 51 as the sum of two consecutive positive integers.
2
n − 1 n + 1
2
2
Solution: Since, n = + , so we can write
2
2 2
2
2
+
−
51 − 1 51 + 1 2601 1 2601 1
2
51 = + = + = 1300 1301
+
2 2 2 2
2
So, 51 can be expressed as sum of two consecutive positive integers, 1300 and 1301.
Alternate Method: Let the two positive consecutive integers be n and n + 1.
Then, n + (n + 1) = 2601
⇒ 2n + 1 = 2601 or 2n = 2600
So, n = 1300
Therefore, the numbers are 1300, 1301 and 1300 + 1301 = 2601.
Example 16: Observe the following pattern and find the missing number.
2
11 = 121
101 = 10201
2
1001 = ………………
2
2
10001 = 100020001
2
100001 = 10000200001
2
1000001 = …………………………
Solution: In the given pattern, we observe that:
(a) The square on the RHS of each equality has an odd number of digits.

(b) The first and the last digits of each number on the RHS are 1.
(c) The middle digit in each number on the RHS is 2.

(d) The number of zeros between the left-most digit 1 and the middle digit 2, or between
the middle digit 2 and the right-most digit 1, is the same as that of zeros in the number
on the LHS.

2
2
Therefore, 1001 = 1002001 and 1000001 = 1000002000001.
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