Question

Challenge for all

No spamm !!!!

Please. I will mark brainliest​

Answer 1

Answer:

this is a answerchallenge accepted

Answer 2

Answer:

1 × 2 × 3 + 2 × 3 × 4 + ... + n(n + 1)(n + 2) = $\frac{n(n + 1)(n + 2)(n + 3)}{4}$ for all n ∈ N

Step-by-step explanation:

Let A = 1 × 2 × 3 + 2 × 3 × 4 + ... + n(n + 1)(n + 2)

⇒4A = 1 × 2 × 3 × 4 + 2 × 3 × 4 × 4 + ... + 4n(n + 1)(n + 2)

  4A = 1 × 2 × 3 × 4 + 2 × 3 × 4 × (5 - 1) + 3 × 4 × 5 × (6 - 2) + ... + n(n + 1)(n + 2) × [(n + 3) - (n - 1)]

  4A = 1 × 2 × 3 × 4 + 2 × 3 × 4 × 5 - 1 × 2 × 3 × 4 + 3 × 4 × 5 × 6 - 2 × 3 × 4 × 5 + ... + n(n + 1)(n + 2)(n + 3) - n(n + 1)(n + 2)(n - 1)

  4A = (1 × 2 × 3 × 4 - 1 × 2 × 3 × 4) + (2 × 3 × 4 × 5 - 2 × 3 × 4 × 5) + (3 × 4 × 5 × 6 - 3 × 4 × 5 × 6) + ... + n(n + 1)(n + 2)(n + 3)

  4A = 0 + 0 + 0 + ... + n(n + 1)(n + 2)(n + 3) = n(n + 1)(n + 2)(n + 3)

  ⇒4A ÷ 4 = A = $\frac{n(n + 1)(n + 2)(n + 3)}{4}$ for all n ∈ N