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  • Find the sum of the series (3^3 – 2^3) + (5^3 – 4^3) + (7^3 – 6^3) + … to (i) n terms

Find the sum of the series (3\textsuperscript{3} - 2\textsuperscript{3}) + (5\textsuperscript{3} - 4\textsuperscript{3}) + (7\textsuperscript{3} - 6\textsuperscript{3}) + \ldots   to

(i) n terms

(ii) 10 terms

Answers (1)

Generalising the series in terms of i

\\ S= \sum _{i=1}^{n} \left[ \left( 2i+1 \right) ^{3}- \left( 2i \right) ^{3} \right] \\\\ = \sum _{i=1}^{n} \left[ \left( 2i+1-2i \right) \left( \left( 2i+1 \right) ^{2}+ \left( 2i+1 \right) \left( 2i \right) + \left( 2i \right) ^{2} \right) \right] \\\\ = \sum _{i=1}^{n} \left[ 4i^{2}+4i+1+4i^{2}+2i+4i^{2} \right] = \sum _{i=1}^{n} \left[ 12i^{2}+6i+1 \right] \\\\

\\ =12 \left[ \frac{n \left( n+1 \right) \left( 2n+1 \right) }{6} \right] +6 \left[ \frac{n \left( n+1 \right) }{2} \right] + \left[ n \right] \\\\ =2 \left[ 2n^{3}+3n^{2}+n \right] +3 \left[ n^{2}+n \right] + \left[ n \right] \\\\

    Sum up to n terms =4n^{3}+9n^{2}+6n \\\\

\\ \text{Sum up to 10 terms }S_{10}=4 \left( 10 \right) ^{3}+9 \left( 10 \right) ^{2}+6 \left( 10 \right) =4960 \\\\

 

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infoexpert21

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