Question

a sequences of term is defined by\({a_1} = 1an = \left( {5 - n} \right){a_{n - 1}}\).

calculate\(\sum\limits_{n = 1}^\infty {an} \).

Short answer

\(\sum\limits_{n = 1}^\infty {an} \)=16

Step-by-step solution

given and to find

\({a_1} = 1\) \(an = {\left( {5 - n} \right)_{an - 1}}\) (given)

\(\sum\limits_{n = 1}^\infty {an} \)(to find)

find values

\({a_1} = 1\)

\({a_2} = \left( {5 - 2} \right){a_{2 - 1}}\) \({a_3} = \left( {5 - 3} \right){a_{3 - 1}}\) \({a_4} = \left( {5 - 4} \right){a_{4 - 1}}\)

= \(\left( 3 \right){a_1}\) =\(\left( 2 \right){a_2}\) =\(\left( 1 \right){a_3}\)

=3x1 = 2x3 =1x6

=3 =6 =6

\({a_5} = \left( {5 - 5} \right){a_{5 - 1}}\) \({a_6} = \left( {5 - 6} \right){a_{6 - 1}}\) \({a_7} = \left( {5 - 7} \right){a_{7 - 1}}\)

\( = \left( 0 \right){a_4}\) = \(\left( { - 1} \right){a_5}\) =\(\left( 2 \right){a_6}\)

\(\begin{aligned}\\\end{aligned}\) =0x6 =-1x0 =2x0

= 0 =0 =0

All from \({a_5}\)will be 0 as \({a_5},{a_6},{a_7}\)all are 0

find the summation

\(\sum\limits_{n = 1}^\infty {an} \)=\(\sum\limits_{n = 1}^4 {an} \)+\(\sum\limits_{n = 1}^\infty {an} \)

=\(\sum\limits_{n = 1}^4 {an} \)+0

=\(\sum\limits_{n = 1}^4 {an} \)

=\({9_1} + {9_2} + {9_3} + {9_4}\)

=1+3+6+6

=16

Hence, \(\sum\limits_{n = 1}^\infty {an} \)=16