Question ID: #906
Let $z=(1+i)(1+2i)(1+3i)….(1+ni)$, where $i =\sqrt{-1}$. If $|z|^{2}=44200$, then $n$ is equal to
Solution:
$$|z|^2 = |(1+i)(1+2i)\dots(1+ni)|^2$$
$$|z|^2 = |1+i|^2 |1+2i|^2 \dots |1+ni|^2$$
$$|z|^2 = (1^2+1^2)(1^2+2^2)\dots(1^2+n^2)$$
$$|z|^2 = (2)(5)(10)(17)(26)\dots(1+n^2)$$
Given $|z|^2 = 44200$.
Factorize 44200:
$$44200 = 100 \times 442 = 100 \times 2 \times 221 = 100 \times 2 \times 13 \times 17$$
$$= (2 \cdot 5 \cdot 2 \cdot 5) \cdot 2 \cdot 13 \cdot 17$$
$$= 2 \cdot 5 \cdot 10 \cdot 17 \cdot 26$$
Comparing the series product:
$n=1 \rightarrow 2$
$n=2 \rightarrow 2 \cdot 5$
$n=3 \rightarrow 2 \cdot 5 \cdot 10$
$n=4 \rightarrow 2 \cdot 5 \cdot 10 \cdot 17$
$n=5 \rightarrow 2 \cdot 5 \cdot 10 \cdot 17 \cdot 26$
This matches the given value.
So, $n = 5$.
Ans. (5)
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