Solution : \(K(10)^9\) = \((10)^9\) + \(2(11)^1(10)^8\) + \(3(11)^2(10)^7\) + …… + \(10(11)^9\) K = 1 + 2\(({11\over 10})\) + 3\(({11\over 10})^2\) + ….. + 10\(({11\over 10})^9\) ……(i) \(({11\over 10})\)K = 1\(({11\over 10})\) + 2\(({11\over 10})^2\) + 3\(({11\over 10})^3\) + ….. + 10\(({11\over 10})^{10}\) …..(ii) On subtracting equation (ii) from (i), …
Solution : By using method of differences, 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)} = \(1\over 8\)(\(V_n\) – \(V_{n-1}\)) \(S_n\) = \({\sum}_{r=1}^{n} T_n\) = \(1\over 8\)(\(V_n\) – \(V_0\)) \(\therefore\) \(S_n\) = \((2n-1)(2n+1)(2n+3)(2n+5)\over 8\) + \(15\over 8\) = \(n(2n^3 + 8n^2 + 7n – 2)\) Similar Questions Find the …
Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + …… Read More »
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 \(V_n\) = \(1\over n(n+1)\) \(\therefore\) …
