Chapter 6: Problem 467
The product of three consecutive natural numbers is divisible by (a) 6 (b) 5 (c) 7 (d) 4
Chapter 6: Problem 467
The product of three consecutive natural numbers is divisible by (a) 6 (b) 5 (c) 7 (d) 4
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Get started for freeLet \(\mathrm{P}(\mathrm{n}): \mathrm{n}^{2}+1\) is an odd integer, if it is assumed that \(\mathrm{P}(\mathrm{k})\) is true \(\Rightarrow \mathrm{P}(\mathrm{k}+1)\) is true. Therefore, \(\mathrm{P}(\mathrm{n})\) is true (a) for \(\mathrm{n}>1\) (b) for all \(\mathrm{n} \in \mathrm{N}\) (c) for \(\mathrm{n}>2\) (d) None of these
If \(\mathrm{P}(\mathrm{n}): 1+3+5+\ldots \ldots+(2 \mathrm{n}-1)=\mathrm{n}^{2}\) is (a) True for \(\mathrm{n}>1\) (b) True for all \(\mathrm{n} \in \mathrm{N}\) (c) True for no \(\mathrm{n}\) (d) None of these
If \(n \in N\), then \(11^{n+2}+12^{2 n+1}\) is divisible by (a) 113 (b) 123 (c) 133 (d) None of these
The inequality \(\mathrm{n} !>2^{\mathrm{n}-1}\) is true for (a) \(n>2, n \in N\) (b) \(n<2\) (c) \(\mathrm{n} \in \mathrm{N}\) (d) None of these
\(x\left(x^{n-1}-n \alpha^{n-1}\right)+\alpha^{n}(n-1)\) is divisible by \((x-\alpha)^{2}\) for (a) \(n>1\) (b) \(n>2\) (c) For all n \(\in \mathrm{N}\) (d) None of these
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