Question

Prove that 2^n + 6x9^n is always divisible by 7 for any positive integer n.​

Answer 1

Answer:

2ⁿ + 6*9ⁿ Divisible by 7

Step-by-step explanation:

Prove that 2^n + 6x9^n is always divisible by 7 for any positive integer n

2ⁿ + 6*9ⁿ

= 2ⁿ + 6*(7+2)ⁿ

(a + b)ⁿ = aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² +........................ +ⁿCn₋₁abⁿ⁻¹ + bⁿ

= 2ⁿ + 6*(7ⁿ + ⁿC₁7ⁿ⁻¹*2 + ⁿC₂aⁿ⁻²b² +........................ +ⁿCn₋₁72ⁿ⁻¹ + 2ⁿ)

= 2ⁿ + 6*2ⁿ + 6*(7ⁿ + ⁿC₁7ⁿ⁻¹*2 + ⁿC₂aⁿ⁻²b² +........................ +ⁿCn₋₁7*2ⁿ⁻¹)

= 2ⁿ(1 + 6) + 6*(7ⁿ + ⁿC₁7ⁿ⁻¹*2 + ⁿC₂aⁿ⁻²b² +........................ +ⁿCn₋₁7*2ⁿ⁻¹)

= 7*2ⁿ + 7*6(7ⁿ⁻¹ + ⁿC₁7ⁿ⁻²*2 + ⁿC₂aⁿ⁻³b² +........................ +ⁿCn₋₁2ⁿ⁻¹)

= 7 *( 2ⁿ + 6(7ⁿ⁻¹ + ⁿC₁7ⁿ⁻²*2 + ⁿC₂aⁿ⁻³b² +........................ +ⁿCn₋₁2ⁿ⁻¹))

Hence 2ⁿ + 6*9ⁿ Divisible by 7

Answer 2

Such type of questions can also solve by Principal of Mathematical Induction.

To Prove: Prove that
$p(n) = {2}^{n} + 6 \times {9}^{n}$
is always divisible by 7 for any positive integer n.

Solution:

first check is the expression true for
n= 1,
put in the given expression

$p(1) = {2}^{1} + 6 \times {9}^{1} \\ \\ = 2 + 54 \\ \\ = 56 \\ \\ 56 = 7 \times 8 \: \: \: \: \: \: 8 \: \epsilon \: Z^+$
Now let us assume that this is also true for n= k

So,

$p(k) = {2}^{k} + 6 \times {9}^{k} \\ \\ {2}^{k} + 6 \times {9}^{k} = 7r \: \: \: \: \: here \: r \: \epsilon \: Z^+ \: \: \: ...eq1$
Now we have to prove that it is also true for n= k+1

put n= k+1

$p(k + 1) = {2}^{k + 1} + 6 \times {9}^{k + 1}.....eq2$
Now try to manipulate eq 2 as eq1,so that we can substitute the value 7r in eq2

For that,look only LHS if eq2 and 3

$p(k + 1) = {2}^k.2 + 6 \times {9}^{k + 1} \\ \\ = 2({2}^k + 6. {9}^{k} - 6. {9}^{k} ) + 6 \times {9}^{k + 1} \\ \\ = 2(7r - 6. {9}^{k} ) + 6. {9}^{k + 1} \\ \\ = 14r - 12. {9}^{k} + 6. {9}^{k} . {9}^{1} \\ \\ = 14r - 12. {9}^{k} + 54. {9}^{k} \\ \\ = 14r + 42. {9}^{k} \\ \\ = 7(2r + 6. {9}^{k} ) \\ \\ p(k + 1)= 7m \: \: \: \: \:here \: \: m=2r + 6. {9}^{k} \\ \\ \\ hence \: p(k + 1) \: is \: also \: divisible \: by \: \\\\7 \: for \: some \: integer \: m$

Hope it helps you.