Q62 / 150 · MHT CET 2026 (16 April Shift 1) · Mathematics
If $(1 + i) \cdot (1 + 2i) \ldots\ldots\ldots (1 + ni) = x + iy$ (Where $i = \sqrt{-1}$ ), then the value of $(2) \cdot (5) \cdot (10)\ldots\ldots\ldots(1 + n^2)$
- A$\dfrac{\sqrt{x} + \sqrt{y}}{2}$
- Correct answerB$x^2 + y^2$
- C$\dfrac{\sqrt{x} + \sqrt{y}}{\sqrt{x} - \sqrt{y}}$
- D$x^2 - y^2$
Solution
Given $(1 + i)(1 + 2i) \dots (1 + ni) = x + iy$
Taking the conjugate on both sides:
$(1 - i)(1 - 2i) \dots (1 - ni) = x - iy$
Multiplying the two equations:
$((1 + i)(1 - i))((1 + 2i)(1 - 2i)) \dots ((1 + ni)(1 - ni)) = (x + iy)(x - iy)$
$(1^2 - i^2)(1^2 - (2i)^2) \dots (1^2 - (ni)^2) = x^2 - (iy)^2$
Since $i^2 = -1$, we get:
$(1 + 1)(1 + 4) \dots (1 + n^2) = x^2 + y^2$
$2 \cdot 5 \cdot 10 \dots (1 + n^2) = x^2 + y^2$
Answer: $x^2 + y^2$
Taking the conjugate on both sides:
$(1 - i)(1 - 2i) \dots (1 - ni) = x - iy$
Multiplying the two equations:
$((1 + i)(1 - i))((1 + 2i)(1 - 2i)) \dots ((1 + ni)(1 - ni)) = (x + iy)(x - iy)$
$(1^2 - i^2)(1^2 - (2i)^2) \dots (1^2 - (ni)^2) = x^2 - (iy)^2$
Since $i^2 = -1$, we get:
$(1 + 1)(1 + 4) \dots (1 + n^2) = x^2 + y^2$
$2 \cdot 5 \cdot 10 \dots (1 + n^2) = x^2 + y^2$
Answer: $x^2 + y^2$
Hence, the correct answer is option: 'B'.