Example 1 : <\/span> If \\({\\sum}_{r=1}^{n\u200e} T_r\\) = \\(n\\over 8\\) (n + 1)(n + 2)(n + 3), then find \\({\\sum}_{r=1}^{n\u200e} \\)\\(1\\over T_r\\)<\/p>\n Solution : <\/span> \\(\\because\\) \\(T_n\\) = \\(S_n – S_{n-1}\\) Example 2 : <\/span> Example 2: Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + ……<\/p>\n Solution : <\/span>The \\(n^{th}\\) term is (2n-1)(2n+1)(2n+3) Practice these given sequences and series examples to test your knowledge on concepts of sequences and series.<\/p>\n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":" Here you will learn some sequences and series examples for better understanding of sequences and series concepts. Example 1 : If \\({\\sum}_{r=1}^{n\u200e} T_r\\) = \\(n\\over 8\\) (n + 1)(n + 2)(n + 3), then find \\({\\sum}_{r=1}^{n\u200e} \\)\\(1\\over T_r\\) Solution : \\(\\because\\) \\(T_n\\) = \\(S_n – S_{n-1}\\) = \\({\\sum}_{r=1}^{n\u200e} T_r\\) – \\({\\sum}_{r=1}^{n\u200e …<\/p>\n
\n\t\t\t = \\({\\sum}_{r=1}^{n\u200e} T_r\\) – \\({\\sum}_{r=1}^{n\u200e – 1} T_r\\)
\n\t\t\t = \\(n(n+1)(n+2)(n+3)\\over 8\\) – \\((n-1)(n)(n+1)(n+2)\\over 8\\)
\n\t\t\t = \\(n(n+1)(n+2)\\over 8\\)[(n+3) – (n-1)] = \\(n(n+1)(n+2)\\over 8\\)(4)
\n\t\t\t \\(T_n\\) = \\(n(n+1)(n+2)\\over 2\\)
\n\t\t\t \\(\\implies\\) \\(1\\over T_n\\) = \\(2\\over n(n+1)(n+2)\\) = \\((n+2)-n\\over n(n+1)(n+2)\\)
\n\t\t\t = \\(1\\over n(n+1)\\) – \\(1\\over (n+1)(n+2)\\)
\n\t\t\t Let \\(V_n\\) = \\(1\\over n(n+1)\\)
\n\t\t\t \\(\\therefore\\) \\(1\\over T_n\\) = \\(V_n\\) – \\(V_{n+1}\\)
\n\t\t\t Putting n = 1, 2, 3, …… n
\n\t\t\t \\(\\implies\\) \\(1\\over T_1\\) + \\(1\\over T_2\\) + \\(1\\over T_3\\) + ….. + \\(1\\over T_n\\) = \\(V_1\\) – \\(V_{n+1}\\)
\n\t\t\t = \\({\\sum}_{r=1}^{n\u200e} \\)\\(1\\over T_r\\) = \\(n^2 + 3n\\over 2(n+1)(n+2)\\)\n\t\t\t<\/p>\n
\n\n
\n\t\t \\(T_n\\) = (2n-1)(2n+1)(2n+3)
\n\t\t \\(T_n\\) = \\(1\\over 8\\)(2n-1)(2n+1)(2n+3){(2n+5) – (2n-3)}
\n\t\t = \\(1\\over 8\\)(\\(V_n\\) – \\(V_{n-1}\\))
\n\t\t \\(S_n\\) = \\({\\sum}_{r=1}^{n\u200e} T_n\\) = \\(1\\over 8\\)(\\(V_n\\) – \\(V_0\\))
\n\t\t \\(\\therefore\\) \\(S_n\\) = \\((2n-1)(2n+1)(2n+3)(2n+5)\\over 8\\) + \\(15\\over 8\\)
\n\t\t = \\(n(2n^3 + 8n^2 + 7n – 2)\\)\n
\n <\/p>