3636
Suppose n is a natural number such that
i + 2i^2 + 3i^3 +................+ ni^n=18√2,where i is √-1.Then n is...
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3 Answers
1057Firstly,the co-efficient of i and the real part will both be integers...so if the mod is 18√2 then surely both parts'(real and imaginary) co-efficients are 18 itself....therefore...
co-efficient of i is (1-3+5-7+9....)..
the above sum upto n terms is equal to (-1)n+1n...
therefore n here is 18..
now every 2 terms out of 4 are imaginary and the other two real...
therefore total no of terms(n to be calculated) is 36.
1133in can either be 1, -1 or i. The left side will be a natural number or a complex number, whereas the right side is irrational. Therefore there is no solution for n.
- Anurag Ghosh Yaa Sourish i forgot to mention dat d sum i + 2i^2 + 3i^3 +................+ ni^n is in a mod....now find for n..Upvote·0· Reply ·2013-02-06 09:10:42