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Q.
The sum $S_{n} = n^{3} + 3n^{2}+ 5n + 3$ is divisible by
Principle of Mathematical Induction
Solution:
Let $P\left(n\right)$ be the statement given by
$P\left(n\right): S_{n} = n^{3} + 3n^{2} + 5n + 3$ is divisible by $3$
We have, $P\left(1\right): S_{1} = 1^{3} + 3\left(1\right)^{2 }+ 5\left(1\right) + 3 = 12$, which is divisible by $3$
$\therefore P\left(1\right)$ is true.
Let $P\left(k\right)$ be true. Then,
$ P\left(k\right): S_{k} = k^{3} + 3k^{2} + 5k + 3$ is divisible by $3$
$\Rightarrow S_{k} =k^{3} + 3 k^{2} + 5k + 3 = 3\lambda$, for some $\lambda\in N\quad...\left(i\right)$
We now wish to show that $P\left(k + 1\right)$ is true, i.e.
$\left(k + 1\right)^{3} + 3\left(k + 1\right)^{2} + 5\left(k + 1 \right)+ 3$ is divisible by $3$
Now, $\left(k + 1\right)^{3} + 3\left(k + 1\right)^{2} + 5\left(k + 1\right) + 3$
$ = \left(k^{3} + 3 k^{2} + 5k + 3\right) + 3k^{2} + 9k + 9$
$ = 3\lambda + 3\left(k^{2} + 3k + 3\right)\quad$ [Using $\left(i\right)$]
$= 3\left[\lambda+ k^{2} + 3k + 3\right]$, which is divisible by $3$
Thus, $P\left(k + 1\right)$ is true, whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction, the statement is true for all $n\in N$.