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 freeFor each \(n \in N, 3^{2 n}-1\) is divisible by (a) 8 (b) 16 (c) 32 (d) None of these
Let \(\mathrm{P}(\mathrm{n}): \mathrm{n}^{2}+1\) 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
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
If \(\mathrm{P}(\mathrm{n}):\left[4^{\mathrm{n}} /(\mathrm{n}+1)\right]<\left[(2 \mathrm{n}) ! /(\mathrm{n} !)^{2}\right]\), then \(\mathrm{P}(\mathrm{n})\) is true for (a) \(\mathrm{n} \geq 1\) (b) \(n>0\) (c) \(n<0\) (d) \(n \geq 2, n \in N\)
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
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