Question Number 217895 by hardmath last updated on 23/Mar/25

Commented by mr W last updated on 23/Mar/25

Commented by hardmath last updated on 23/Mar/25

Commented by mr W last updated on 23/Mar/25

Commented by Frix last updated on 23/Mar/25

Commented by mr W last updated on 23/Mar/25

Commented by hardmath last updated on 23/Mar/25
![Using (-1)ᵏ = i²ᵏ and (-1)ᵏ⁺¹ = i²ᵏ⁺²: [∑ₖ[1→+∞] (-1)ᵏ C(2k, n)]² + [∑ₖ[1→+∞] (-1)ᵏ⁺¹ C(2k-1, n)]² = [∑ₖ[1→+∞] i²ᵏ C(2k, n)]² + i³.[∑ₖ[1→+∞] i²ᵏ⁻¹ C(2k-1, n)]² = [Re{(1+i)ⁿ}]² + i⁴ [Im{(1+i)ⁿ}]² = |(1+i)ⁿ|² = |1+i|²ⁿ = (√2)²ⁿ = 2ⁿ](https://www.tinkutara.com/question/Q217970.png)
[∑ₖ[1→+∞] (-1)ᵏ C(2k, n)]²
+ [∑ₖ[1→+∞] (-1)ᵏ⁺¹ C(2k-1, n)]²
= [∑ₖ[1→+∞] i²ᵏ C(2k, n)]²
+ i³.[∑ₖ[1→+∞] i²ᵏ⁻¹ C(2k-1, n)]²
= [Re{(1+i)ⁿ}]² + i⁴ [Im{(1+i)ⁿ}]²
= |(1+i)ⁿ|²
= |1+i|²ⁿ
= (√2)²ⁿ
= 2ⁿ
Answered by SdC355 last updated on 23/Mar/25
