Sequences and Series Examples

SEQUENCES & SERIES EXAMPLES

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)}

= \(1\over 8\)(\(V_n\) - \(V_{n-1}\))

\(S_n\) = \({\sum}_{r=1}^{n‎} T_n\) = \(1\over 8\)(\(V_n\) - \(V_0\))

\(\therefore\)     \(S_n\) = \((2n-1)(2n+1)(2n+3)(2n+5)\over 8\) + \(15\over 8\)

= \(n(2n^3 + 8n^2 + 7n - 2)\)

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