Question

(n+1/2n) and (n²+1/4n²-1/2)​

Answer 1

Step-by-step explanation:

N is odd integers .

so, Let n = 2k + 1 , k ∈ ℕ

now, n⁴ + 4n² + 11

= (2k +1)² + 4(2k + 1)² + 11

= (4k² + 4k + 1)² + 4(4k² + 4k + 1) + 11

= (16k⁴ + 16k² + 1 + 32k³ + 8k + 8k²) + 16k² + 16k + 4 + 11

= 16k⁴ + 32k³ + 40k² + 24k + 16

= 16[k⁴ + 2k³ + 1] + 40k² + 24k

= 16[k⁴ + 2k² + 1] + (80k² + 48k)/2

= 16[k⁴ + 2k² + 1] + 16k(5k + 3)/2

here, k(5k + 3)/2 is also a integer for all natural number of k

so, assume k(5k + 3)/2 = m

now, n⁴ + 4n² + 11

= 16[k⁴ + 2k² + 1] + 16m

= 16[k⁴ + 2k² + 1 + m]

hence, it is clear that 16 divides n⁴ + 4n² + 11 when n is odd integers.

Answer 2

Answer:

Solution:

Let the given statement be P(n). Then,

P(n): 1² + 2² + 3² + ..... +n² = (1/6){n(n + 1)(2n + 1)}.

Putting n =1 in the given statement, we get

LHS = 1² = 1 and RHS = (1/6) × 1 × 2 × (2 × 1 + 1) = 1.

Therefore LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): 1² + 2² + 3² + ..... + k² = (1/6){k(k + 1)(2k + 1)}.

Now, 1² + 2² + 3² + ......... + k² + (k + 1)²

= (1/6) {k(k + 1)(2k + 1) + (k + 1)²

= (1/6){(k + 1).(k(2k + 1)+6(k + 1))}

= (1/6){(k + 1)(2k² + 7k + 6})

= (1/6){(k + 1)(k + 2)(2k + 3)}

= 1/6{(k + 1)(k + 1 + 1)[2(k + 1) + 1]}

⇒ P(k + 1): 1² + 2² + 3² + ….. + k² + (k+1)²

= (1/6){(k + 1)(k + 1 + 1)[2(k + 1) + 1]}

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