Sequences and 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$ = $n^2 + 3n\over 2(n+1)(n+2)$

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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.