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The
sum of first 9 terms of the series $$\frac{{{1}^{{3}}}}{{1}}+\frac{{{1}^{{3}}+{2}^{{3}}}}{{{1}+{3}}}+\frac{{{1}^{{3}}+{2}^{{3}}+{3}^{{3}}}}{{{1}+{2}+{3}}}+.....$$
is
(1)
71 (2) 96 (3) 142 (4) 192  Speed
00:00
04:32 QuestionMathsClass 12

The
sum of first 9 terms of the series $$\frac{{{1}^{{3}}}}{{1}}+\frac{{{1}^{{3}}+{2}^{{3}}}}{{{1}+{3}}}+\frac{{{1}^{{3}}+{2}^{{3}}+{3}^{{3}}}}{{{1}+{2}+{3}}}+.....$$
is
(1)
71 (2) 96 (3) 142 (4) 192

$${T}_{{n}}=\frac{{{1}^{{3}}+{2}^{{3}}+{3}^{{3}}+\ldots\ldots+{n}^{{3}}}}{{{1}+{3}+{5}+{7}+\ldots\ldots+{\left({2}{n}-{1}\right)}}}$$

$$=\frac{{\left(\frac{{{n}{\left({n}+{1}\right)}}}{{2}}\right)}^{{2}}}{{\frac{{n}}{{2}}{\left[{1}+{\left({2}{n}-{1}\right)}\right]}}}$$

$$={n}^{{2}}\frac{{\left({n}+{1}\right)}^{{2}}}{{{4}\times{n}^{{2}}}}$$

$${T}_{{n}}=\frac{{\left({n}+{1}\right)}^{{2}}}{{4}}$$

$${\sum_{{{x}={1}}}^{{9}}}{T}_{{n}}={\sum_{{{n}={1}}}^{{9}}}\frac{{\left({n}+{1}\right)}^{{2}}}{{4}}$$

$$=\frac{{1}}{{4}}{\sum_{{{n}={1}}}^{{9}}}{\left({n}+{1}\right)}^{{2}}$$

= $$\frac{{1}}{{4}}{\left({2}^{{2}}+{3}^{{2}}+{4}^{{2}}+\ldots+{10}^{{2}}\right)}$$

$$=\frac{{1}}{{4}}{\left({1}^{{2}}+{2}^{{2}}+{3}^{{2}}+\ldots.+{10}^{{2}}-{1}^{{2}}\right)}$$

$$=\frac{{1}}{{4}}{\left[\frac{{{10}{\left({10}+{1}\right)}{\left({2}\times{10}+{1}\right)}}}{{6}}-{1}\right]}$$

$$=\frac{{1}}{{4}}{\left[\frac{{{10}\times{11}\times{27}}}{{6}}-{1}\right]}$$

$$=\frac{{1}}{{4}}\times{384}={96}$$

option 1 is correct          