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If (1+i)(1+2i)(1+3i)...(1+n i)=a+i b , ...

If `(1+i)(1+2i)(1+3i)...(1+n i)=a+i b ,`
then `2. 5. 10. 17... (1+n^2\ )=`
a. `a-i b`
b. `a^2-b^2`
c. `a^2+b^2`
d. none of these

Text Solution

Verified by Experts

Given that `(1+i)(1+2i)(1+3i)...(1+n i)=a+i b ,`
If we take modulus on both sides, `|axxb|=|a|xx|b|`
`=>sqrt((1+1)(1+2^2)....(1+n^2))=sqrt(a^2+b^2)`
Squaring both sides we get `2. 5. 10. 17... (1+n^2\ )=``a^2+b^2`
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