PUMPA - SMART LEARNING
எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்
Book Free DemoNumbers between square numbers:
Numbers between \(1^2\) and \(2^2\) are \(2\), \(3\).
So, there are \(2 = 2(1)\) numbers between \(1^2\) and \(2^2\).
Numbers between \(2^2\) and \(3^2\) are \(5\), \(6\), \(7\), \(8\).
So, there are \(4 = 2(2)\) numbers between \(2^2\) and \(3^2\).
Numbers between \(3^2\) and \(4^2\) are \(10\), \(11\), \(12\), \(13\), \(14\), \(15\).
So, there are \(6 = 2(3)\) numbers between \(3^2\) and \(4^2\).
In general, we can say that there are \(2n\) non-perfect square numbers between the squares of the numbers \(n\) and \((n + 1)\).
Summing up odd natural numbers:
Odd numbers are \(1\), \(3\), \(5\), \(7\), \(9\), \(11\), \(13\), \(15\), ...
First odd number \(=\) \(1 = 1^2\)
Sum of first two odd numbers \(=\) \(1 + 3 = 4 = 2^2\)
Sum of first three odd numbers \(=\) \(1 + 3 + 5 = 9 = 3^2\)
Sum of first four odd numbers \(=\) \(1 + 3 + 5 + 7 = 16 = 4^2\)
Sum of first five odd numbers \(=\) \(1 + 3 + 5 + 7 + 9 = 25 = 5^2\)
….
Sum of first \(n\) odd numbers \(=\) \(1 + 3 + 5 + 7 + 9 + 11 + … = n^2\)
The sum of the first \(n\) consecutive odd natural numbers is \(n^2\).
Summing up two consecutive natural numbers results in an odd square:
Let us take any odd natural number, say \(9\).
The square of \(9\) is \(81\).
To find the sum of two consecutive natural numbers equal to the odd square, we can use the below formula.
Now, let's take \(a = 9\).
Substitute the value of \(a\) in the above formula.
\(81 = 40 + 41\)
The numbers \(40\) and \(41\) are consecutive.
Therefore, \(81\) can be written as the sum of \(40\) and \(41\).
The square of any odd number can be written as the sum of two consecutive natural numbers.
Product of two consecutive even or odd natural numbers:
Let us take two consecutive even natural numbers, \(4\) and \(6\).
\(4 \times 6 = 24 = 25 - 1 = 5^2 - 1\)
Here, \(4\) can be written as \(5 - 1\), and \(6\) can be written as \(5 + 1\).
So, \((5 - 1) \times (5 + 1) = 5^2 - 1\).
Now, let us take two consecutive odd natural numbers, \(7\) and \(9\).
\(7 \times 9 = 63 = 64 - 1 = 8^2 - 1\)
Here, \(7\) can be written as \(8 - 1\), and \(9\) can be written as \(8 + 1\).
So, \((8 - 1) \times (8 + 1) = 8^2 - 1\).
In general, we can say that \((a - 1) \times (a + 1) = a^2 - 1\).
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