Question

Show that 3 · 4^n+ 51 is divisible by 3 and 9 for all positive integers n.

Answer 1

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$3 \times {4}^{n} + 51$

$= 3 \times {(1 + 3)}^{n} + 51$

$= 3 \times \{ \: 1 + \binom{n}{1} \times 3 + \binom{n}{2} \times {3}^{2} + \binom{n}{3} \times {3}^{3} + ............ + {3}^{n} \: \} + 51$

$= 3 \times \{ \: 18 + \binom{n}{1} \times 3 + \binom{n}{2} \times {3}^{2} + \binom{n}{3} \times {3}^{3} + ............ + {3}^{n} \: \}$

$= 9\times \{ \: 2 + \binom{n}{1} + \binom{n}{2} \times {3}^{} + \binom{n}{3} \times {3}^{2} + ............ + {3}^{(n - 1)} \: \}$

So $3 \times {4}^{n} + 51$ has a factor 9

Hence $3 \times {4}^{n} + 51$ is divisible by 3 & 9

$</p><p></p><p>\displaystyle\textcolor{red}{Please \: Mark \: it \: Brainliest}</p><p>$

Answer 2

✪ANSWER✪

The given expression is

$\:\:\:\:\: 3.4^n + 51$

$= 3(4^n + 17)$

As, it is clear that 3 is a factor, the given expression is divisible by 3.

Now,

$4^n = (3+1)^n$

Using Binomial expansion,

$4^n = \sum \limits_{r=0}^{n}(^nC_r3^r)$

$4^n = \sum \limits_{r=1}^{n}(^nC_r3^r) + (^nC_03^0)$

$4^n = 3\sum \limits_{r=1}^{n}(^nC_r3^{r-1} ) + 1$

$4^n = 3q + 1$

where q is an integer.

[ This can also be concluded by writing out values of $4^n$ for various values of n]

So,

$\:\:\:\:\:3(4^n + 17)$

$= 3(3q+1 + 17)$

$= 3(3q + 18)$

$= 3\times 3(q + 6)$

$= 9(q + 6)$

Since 9 is a factor, the given expression is divisible by 9.

Hence Proved