Question

Using the principle of mathematical induction prove that for all n belongs toN:
2 + 4 + 6 + ... + 2n
... + 2n = n² + n.​

Answer 1

★​To prove:-

For all n belongs to N,

2 + 4 + 6 + ...+ 2n = n² + n.​

★Proof:-

Let,

• f(n) :2 + 4 + 6+ …+2n = n²+ n

Then f(l) will be:

f(l): 2 = l² + 1

         = 2

Therefore,

f(l) is true

Let f(n) is true for some natural number n = p.

Then,

→f(p): 2 + 4 + 6 +...+2p = p² + p---(1)

Now,

f(p + l): 2 + 4 + 6 + 8+ …+2p+ 2 (p+1)

From equation (1),

➜ p² + p + 2(p+ 1)  

➜ p² + p + 2p + 2

➜ p² + 2p+1+p+1

➜ (p+1)² + p+1

Hence,

We can see that f(p + 1) is true whenever f(p) is true.

Hence,

By the principle of mathematical induction, for any natural number n,

2 + 4 + 6 + ... +2n = n² + n.​

______________

Answer 2

Question:

Using the principle of mathematical induction prove that for all n belongs toN:

2 + 4 + 6 + ... + 2n

... + 2n = n² + n.

WE HAVE TO PROOVE:

For all n belongs to N, 2 + 4 + 6 + ...+ 2n = n² + n.

Proof:

➪Let

f(n) :2 + 4 + 6+ …+2n = n²+ n

Then f(l) will be:

f(l): 2 = l² + 1

         = 2

Therefore,

f(l) is true

Let f(n) is true for some natural number n = p.

Then,

f(p): 2 + 4 + 6 +...+2p = p² + p---(1)

Now,

f(p + l): 2 + 4 + 6 + 8+ …+2p+ 2 (p+1)

From equation (1),

p² + p + 2(p+ 1)  

p² + p + 2p + 2

p² + 2p+1+p+1

(p+1)² + p+1

Hence,

We can see that f(p + 1) is true whenever f(p) is true.

Therefore

By the principle of mathematical induction, for any natural number n,

2 + 4 + 6 + ... +2n = n² + n.

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