{"id":6871,"date":"2021-10-21T21:28:58","date_gmt":"2021-10-21T15:58:58","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6871"},"modified":"2021-10-25T02:26:05","modified_gmt":"2021-10-24T20:56:05","slug":"if-sum_r1n%e2%80%8e-t_r-nover-8-n-1n-2n-3-then-find-sum_r1n%e2%80%8e-1over-t_r","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/if-sum_r1n%e2%80%8e-t_r-nover-8-n-1n-2n-3-then-find-sum_r1n%e2%80%8e-1over-t_r\/","title":{"rendered":"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\\)"},"content":{"rendered":"
\\(\\because\\) \\(T_n\\) = \\(S_n – S_{n-1}\\)<\/p>\n
= \\({\\sum}_{r=1}^{n\u200e} T_r\\) – \\({\\sum}_{r=1}^{n\u200e – 1} T_r\\)<\/p>\n
= \\(n(n+1)(n+2)(n+3)\\over 8\\) – \\((n-1)(n)(n+1)(n+2)\\over 8\\)<\/p>\n
= \\(n(n+1)(n+2)\\over 8\\)[(n+3) – (n-1)] = \\(n(n+1)(n+2)\\over 8\\)(4)<\/p>\n
\\(T_n\\) = \\(n(n+1)(n+2)\\over 2\\)<\/p>\n
\\(\\implies\\) \\(1\\over T_n\\) = \\(2\\over n(n+1)(n+2)\\) = \\((n+2)-n\\over n(n+1)(n+2)\\)<\/p>\n
= \\(1\\over n(n+1)\\) – \\(1\\over (n+1)(n+2)\\)<\/p>\n
Let \\(V_n\\) = \\(1\\over n(n+1)\\)<\/p>\n
\\(\\therefore\\) \\(1\\over T_n\\) = \\(V_n\\) – \\(V_{n+1}\\)<\/p>\n
Putting n = 1, 2, 3, …… n<\/p>\n
\\(\\implies\\) \\(1\\over T_1\\) + \\(1\\over T_2\\) + \\(1\\over T_3\\) + ….. + \\(1\\over T_n\\) = \\(V_1\\) – \\(V_{n+1}\\)<\/p>\n
= \\({\\sum}_{r=1}^{n\u200e} \\)\\(1\\over T_r\\) = \\(n^2 + 3n\\over 2(n+1)(n+2)\\)<\/p>\n
Let \\(a_n\\) be the nth term of an AP. If \\(\\sum_{r=1}^{100}\\) \\(a_{2r}\\) = \\(\\alpha\\) and \\(\\sum_{r=1}^{100}\\) \\(a_{2r-1}\\) = \\(\\beta\\), then the common difference of the AP is<\/a><\/p>\n Prove that the sum of first n natural numbers is \\(n(n+1)\\over 2\\)<\/a><\/p>\n If \\((10)^9\\) + \\(2(11)^1(10)^8\\) + \\(3(11)^2(10)^7\\) + \u2026\u2026 + \\(10(11)^9\\) = \\(K(10)^9\\), then k is equal to<\/a><\/p>\n If x, y and z are in AP and \\(tan^{-1}x\\), \\(tan^{-1}y\\) and \\(tan^{-1}z\\) are also in AP, then<\/a><\/p>\n