Suppose n is a natural number such that |i+2 i^{2}+3 i^{3}+. . .+n i^{n}=18 \sqrt{2}|. where i is the square root of -1. Then n is

KVPY SB SX Exam 2011 Question Paper

Question : 2 of 120

Marks: +1, -0

Suppose n is a natural number such that

|i+2 i^{2}+3 i^{3}+. . .+n i^{n}=18 √{2}|

. where i is the square root of -1. Then n is

Solution:

{S}= {i}+2 {i}^{2}+3 {i}^{3}+. . . . .+ {ni}^{ {n}}

\;{ {iS}= {i}^{2}+2 {i}^{3}+. . . . .+( {n}-1) {i}^{ {n}}+ {ni}^{ {n}+1}}/{ {S}(1- {i})= {i}+ {i}^{2}+ {i}^{3}+. . . . .+ {i}^{ {n}}- {ni}^{ {n}+1}}

{S}(1- {i})=\;{ {i}(1- {i}^{ {n}})}/{1- {i}}- {ni}^{ {n}+1}

⇒ {S}=\;{1- {i}^{ {n}}}/{-2 {i}}↙{↓}↙{ {z}_{1}( say )}-\;{ {ni}^{ {n}+ {i}}}/{1- {i}}↙{↓ }↙{{z}_{2} (say)}

| {z}_{1}|=\;{1}/{√{2}}

0 \;\; | {z}_{2}|=\;{n}/{√{2}}=\;{n}/{2} √{2}

\;{n}/{2}=18 ⇒ {n}=36

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