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TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Section: Mathematics

Question : 8 of 160

Marks: +1, -0

\omega

\sum\limits_{k=1}^{n}\left(k+\frac{1}{\omega}\right)\left(k+\frac{1}{\omega^{2}}\right)

340

n

Solution:

\omega

is cube root of unity.

\therefore 1+\omega+\omega^{2}=0

and

\omega^{3}=1

\Rightarrow\;\frac{1}{\omega}=\omega^{2}

and

\;\frac{1}{\omega^{2}}=\omega

\omega-\infty\omega^{2}

Now,

\left(k+\;\frac{1}{\omega}\right)\left(k+\;\frac{1}{\omega^{2}}\right)=(k+\omega^{2})(k+\omega)

=k^{2}+(\omega+\omega^{2})k+\omega^{3}

=k^{2}-k+1

\because\;\;\sum\limits_{k=1}^{n}\left(k+\;\frac{1}{\omega}\right)\left(k+\;\frac{1}{\omega^{2}}\right)=340

\Rightarrow\sum\limits_{k=1}^{n}(k+\omega^{2})(k+\omega)=340

\Rightarrow\sum\limits_{k=1}^{n}[k^{2}+(\omega^{2}+\omega)k+\omega^{3}]=340

\Rightarrow\sum\limits_{k=1}^{n}(k^{2}-k+1)=340

\Rightarrow\sum\limits_{k=1}^{n}k^{2}-\sum\limits_{k=1}^{n}k+n=340

\Rightarrow\;\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+n=340

\Rightarrow\;\frac{n(n+1)(2n+1)-3n(n+1)+6n}{6}=340

\Rightarrow\;\;\frac{2n^{3}+4n}{6}=340

\Rightarrow\;\;n(n^{2}+2)=1020

\Rightarrow\;\;n(n^{2}+2)=10(10^{2}+2)

\Rightarrow\;\;n=10

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