Question

prove that 2n +6 x9n is always divisiable by 7for any postive integers
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Answer 1

Answer:

Refer The Above attachment

Answer 2

Given : -

Expression $2^n+6\times 9^n2$

To prove:-

that expression is always divisible by 7 for any positive integer n?

Solution :-

Using Principal of mathematical induction PMI,

Put n=1,

$2^n+6\times 9^n2$

$2^1+6\times 9^12$

2+542+54

5656

56 is divisible by 7.

So, For n=1 it is true.

Now, Assume that for n=k it is true.

So, 2^k+6\times 9^k is divisible by 7.

i.e. $2^k+6\times 9^k=7m$where m is an integer.

To prove for n=k+1

Put n=k+1

$\begin{gathered}=2^{k+1}+6\times 9^{k+1}\\=2^k.2+6(9^k.9)\\=2^k.2+(6.9^k))9\\=2^k.2+(7m-2^k)9\\=2.2^k+63m-9.2^k\\=63m+2.2^k-9.2^k\\=63m-7.2^k\\=7(9m-2^k) \\\end{gathered}$

Which is divisible by 7.

So, For n=k+1 it is true.

Hence by mathematical induction, P(n) is true for all natural number.