Here you will learn some sequences and series examples for better understanding of sequences and series concepts.
Example 1 : If \({\sum}_{r=1}^{n} T_r\) = \(n\over 8\) (n + 1)(n + 2)(n + 3), then find \({\sum}_{r=1}^{n} \)\(1\over T_r\)
Solution : \(\because\) \(T_n\) = \(S_n – S_{n-1}\)
= \({\sum}_{r=1}^{n} T_r\) – \({\sum}_{r=1}^{n – 1} T_r\)
= \(n(n+1)(n+2)(n+3)\over 8\) – \((n-1)(n)(n+1)(n+2)\over 8\)
= \(n(n+1)(n+2)\over 8\)[(n+3) – (n-1)] = \(n(n+1)(n+2)\over 8\)(4)
\(T_n\) = \(n(n+1)(n+2)\over 2\)
\(\implies\) \(1\over T_n\) = \(2\over n(n+1)(n+2)\) = \((n+2)-n\over n(n+1)(n+2)\)
= \(1\over n(n+1)\) – \(1\over (n+1)(n+2)\)
Let \(V_n\) = \(1\over n(n+1)\)
\(\therefore\) \(1\over T_n\) = \(V_n\) – \(V_{n+1}\)
Putting n = 1, 2, 3, …… n
\(\implies\) \(1\over T_1\) + \(1\over T_2\) + \(1\over T_3\) + ….. + \(1\over T_n\) = \(V_1\) – \(V_{n+1}\)
= \({\sum}_{r=1}^{n} \)\(1\over T_r\) = \(n^2 + 3n\over 2(n+1)(n+2)\)
Example 2 : Example 2: Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + ……
Solution : The \(n^{th}\) term is (2n-1)(2n+1)(2n+3)
\(T_n\) = (2n-1)(2n+1)(2n+3)
\(T_n\) = \(1\over 8\)(2n-1)(2n+1)(2n+3){(2n+5) – (2n-3)}
= \(1\over 8\)(\(V_n\) – \(V_{n-1}\))
\(S_n\) = \({\sum}_{r=1}^{n} T_n\) = \(1\over 8\)(\(V_n\) – \(V_0\))
\(\therefore\) \(S_n\) = \((2n-1)(2n+1)(2n+3)(2n+5)\over 8\) + \(15\over 8\)
= \(n(2n^3 + 8n^2 + 7n – 2)\)
Practice these given sequences and series examples to test your knowledge on concepts of sequences and series.