If \(1^{2} + 2^{2} + 2^{2} = 3^{2}\)
\(2^{2} + 3^{2} + 6^{2} = 7^{2}\)
\(3^{2} + 4^{2} + 12^{2} = 13^{2}\)
then \(6^{2} + 7^{2} + 42^{2} = ?\)
TNPSC Group 1 - 2017, TNPSC - 2017,
Answer: C
The answer is \(43^{2}\).
In each line the three numbers are \(n\), \(n + 1\) and their product \(n(n + 1)\), and the right-hand side is the square of (product + 1):
\[ \begin{aligned} 1, 2, 2 &\to 2 + 1 = 3 \\ 2, 3, 6 &\to 6 + 1 = 7 \\ 3, 4, 12 &\to 12 + 1 = 13 \\ 6, 7, 42 &\to 42 + 1 = 43 \end{aligned} \]
Check: 36 + 49 + 1764 = 1849 = \(43^{2}\). \(42^{2}\) forgets the +1, and \(45^{2}\) and \(49^{2}\) do not fit the identity.
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