$$ \frac{2^{3}-1^{3}}{1 \times 7}+\frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 \times 11}+\frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 \times 15}+\cdots+ \frac{30^{3}-29^{3}+28^{3}-27^{3}+\ldots+2^{3}-1^{3}}{15 \times 63}$$ is equal to _____________.
Answer (integer)
120
Solution
<p>$${T_n} = {{\sum\limits_{k = 1}^n {\left[ {{{(2k)}^3} - {{(2k - 1)}^3}} \right]} } \over {n(4n + 3)}}$$</p>
<p>$$ = {{\sum\limits_{k = 1}^n {4{k^2} + {{(2k - 1)}^2} + 2k(2k - 1)} } \over {n(4n + 3)}}$$</p>
<p>$= {{\sum\limits_{k = 1}^n {(12{k^2} - 6k + 1)} } \over {n(4n + 3)}}$</p>
<p>$= {{2n(2{n^2} + 3n + 1) - 3{n^2} - 3n + n} \over {n(4n + 3)}}$</p>
<p>$= {{{n^2}(4n + 3)} \over {n(4n + 3)}} = n$</p>
<p>$\therefore$ ${T_n} = n$</p>
<p>${S_n} = \sum\limits_{n = 1}^{15} {{T_n} = {{15 \times 16} \over 2} = 120}$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Special Series
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