For positive integers n - n 3-2 n is always divisible byA- 3B- 7C- 5D- 6

The correct option is A 3Let, p(n)=n^3+2n ⇒ At n=1, p(1)=1^3+2×1=3 which is divisible by 3 ∴ p(1) is true ⇒ Let p(k) is true; p(k)=k^3+2k=3I ∴ p(k+1)= (k+1)^3 ...

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Standard XII

Mathematics

A.G.P

For positive ...

Question

For positive integers $n, n^{3} + 2 n$ is always divisible by

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Solution

The correct option is A $3$
Let, $p (n) = n^{3} + 2 n$
$\Rightarrow$ At $n = 1, p (1) = 1^{3} + 2 \times 1 = 3$ which is divisible by $3$
$∴ p (1)$ is true
$\Rightarrow$ Let $p (k)$ is true; $p (k) = k^{3} + 2 k = 3 I$
$∴ p (k + 1) = (k + 1)^{3} + 2 (k + 1)$
$= k^{3} + 1 + 3 k^{2} + 3 k + 2 k + 2$
$= k^{3} + 2 k + 3 (k^{2} + k + 1)$
$= 3 I + 3 (k^{2} + k + 1)$
$= 3 I_{1}$
Which is divisible by $3$ .

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For positive integers n,n3+2n is always divisible by

The correct option is A 3Let, p(n)=n3+2n ⇒ At n=1, p(1)=13+2×1=3 which is divisible by 3 ∴p(1) is true ⇒ Let p(k) is true; p(k)=k3+2k=3I ∴p(k+1)=(k+1)3+2(k+1) =k3+1+3k2+3k+2k+2 =k3+2k+3(k2+k+1) =3I+3(k2+k+1) =3I1 Which is divisible by 3.

For positive integers $n, n^{3} + 2 n$ is always divisible by