Question

using mathematical induction prove that 2^3n -1 is divisible by 7
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Answer 1

Step-by-step explanation:

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a. Sodium amalgam, commonly denoted as Na(Hg), it is an alloy of mercury and sodium.

b. Bauxite(Al2O3.2H2O) is the common ore of aluminium.

c. Metal oxides which react with both acids as well as bases to produce salts and water are also known as amphoteric oxides. For example: Al2O3 is an amphoteric oxide. An amphoteric compound is a molecule or ion that can react both an acid as well as a base to produce salt and water.

d. The device used for grinding an ore is grinding mill.

e. Graphite an allotrope of carbon is a good conductor of electricity.

f. Aqua regia is 1:3 mixture of concentrated nitric and hydrochloric acids. It dissolves noble metals such as gold, palladium, and platinum.

Answer 2

To prove :

2^(3n) - 1 is divisible by 7

Proof :

Let P(n) : 2^(3n) - 1 is divisible by 7 be a statement .

If n = 1 , then

2^(3n) - 1 = 2^(3•1) - 1

= 2³ - 1

= 8 - 1

= 7 (which is divisible by 7)

Clearly ,

For n = 1 , 2^(3n) - 1 is divisible by 7 .

Thus ,

P(1) is true .

Now ,

Let us assume that P(k) , k € N is true .

ie .

→ 2^(3k) - 1 is divisible by 7

→ 2^(3k) - 1 = 7m , m € N -----(1)

Now ,

If n = k + 1 , then

→ 2^[3(k+1)] - 1

= 2^(3k + 3) - 1

= 2^3k • 2³ - 1

= (7m + 1)•2³ - 1 [ using eq-(1) ]

= (7m + 1)•8 - 1

= 7•8m + 8 - 1

= 7•8m + 7

= 7•(8m + 1)

= 7p where p = (8m + 1) € N (divisible by 7)

Clearly ,

For n = k + 1 , 2^(3n) - 1 is divisible by 7 .

Thus ,

P(k + 1) is true .

Since ,

P(k + 1) is true whenever P(k) is true , thus

P(n) is true for all n € N .

Hence ,

2^3n - 1 is divisible by 7 .

Hence proved .