Question

. [2^2n] - 1 is divisible by 3.
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Answer 1

Answer:

No... It's incorrect..

Hope it's helpful☺ ❤

Answer 2

Answer:

Let the statement P(n) given as

$\rm \: P(n) : 2^{2n }– 1 \: is \: divisible \: by \: 3,$

for every natural number n.

We observe that P(1) is true, since 2² – 1 = 4 – 1 = 3.1 is divisible by 3.

Assume that P(n) is true for some natural number k, i.e.,

$\rm \: P(k): 2^{ 2k} – 1 \: is \: divisible \: by \: 3,$

i.e., 22k – 1 = 3q, where q ∈ N

Now, to prove that P(k + 1) is true,

we have

$\scriptsize\rm \: P(k + 1) : 2 ^{2(k+1) }– 1 = 2^{2k }+ 2 – 1 = 2^{ 2k} . {2}^{2} – 1$

$\rm = 2^{ 2k} . 4 – 1 = 3. {2}^{2k} + (2^{2k} – 1)$

$= 3.2^{2k }+ 3q$

$\rm = 3 (2^{2k }+ q) = 3m, where \: m ∈ N$

Thus P(k + 1) is true, whenever P(k) is true.

Hence, by the Principle of Mathematical Induction P(n) is true for all natural numbers n

Answer 3

Answer:

Let the statement P(n) given as

$\rm \: P(n) : 2^{2n }– 1 \: is \: divisible \: by \: 3,$

for every natural number n.

We observe that P(1) is true, since 2² – 1 = 4 – 1 = 3.1 is divisible by 3.

Assume that P(n) is true for some natural number k, i.e.,

$\rm \: P(k): 2^{ 2k} – 1 \: is \: divisible \: by \: 3,$

i.e., 22k – 1 = 3q, where q ∈ N

Now, to prove that P(k + 1) is true,

we have

$\scriptsize\rm \: P(k + 1) : 2 ^{2(k+1) }– 1 = 2^{2k }+ 2 – 1 = 2^{ 2k} . {2}^{2} – 1$

$\rm = 2^{ 2k} . 4 – 1 = 3. {2}^{2k} + (2^{2k} – 1)$

$= 3.2^{2k }+ 3q$

$\rm = 3 (2^{2k }+ q) = 3m, where \: m ∈ N$

Thus P(k + 1) is true, whenever P(k) is true.

Hence, by the Principle of Mathematical Induction P(n) is true for all natural numbers n

Answer 4

Answer:

Let the statement P(n) given as

$\rm \: P(n) : 2^{2n }– 1 \: is \: divisible \: by \: 3,$

for every natural number n.

We observe that P(1) is true, since 2² – 1 = 4 – 1 = 3.1 is divisible by 3.

Assume that P(n) is true for some natural number k, i.e.,

$\rm \: P(k): 2^{ 2k} – 1 \: is \: divisible \: by \: 3,$

i.e., 22k – 1 = 3q, where q ∈ N

Now, to prove that P(k + 1) is true,

we have

$\scriptsize\rm \: P(k + 1) : 2 ^{2(k+1) }– 1 = 2^{2k }+ 2 – 1 = 2^{ 2k} . {2}^{2} – 1$

$\rm = 2^{ 2k} . 4 – 1 = 3. {2}^{2k} + (2^{2k} – 1)$

$= 3.2^{2k }+ 3q$

$\rm = 3 (2^{2k }+ q) = 3m, where \: m ∈ N$

Thus P(k + 1) is true, whenever P(k) is true.

Hence, by the Principle of Mathematical Induction P(n) is true for all natural numbers n

Answer 5

Answer:

Let the statement P(n) given as

$\rm \: P(n) : 2^{2n }– 1 \: is \: divisible \: by \: 3,$

for every natural number n.

We observe that P(1) is true, since 2² – 1 = 4 – 1 = 3.1 is divisible by 3.

Assume that P(n) is true for some natural number k, i.e.,

$\rm \: P(k): 2^{ 2k} – 1 \: is \: divisible \: by \: 3,$

i.e., 22k – 1 = 3q, where q ∈ N

Now, to prove that P(k + 1) is true,

we have

$\scriptsize\rm \: P(k + 1) : 2 ^{2(k+1) }– 1 = 2^{2k }+ 2 – 1 = 2^{ 2k} . {2}^{2} – 1$

$\rm = 2^{ 2k} . 4 – 1 = 3. {2}^{2k} + (2^{2k} – 1)$

$= 3.2^{2k }+ 3q$

$\rm = 3 (2^{2k }+ q) = 3m, where \: m ∈ N$

Thus P(k + 1) is true, whenever P(k) is true.

Hence, by the Principle of Mathematical Induction P(n) is true for all natural numbers n