Let S be the sum of the first 9 terms of the series : x + ka + x2 + (k + 2) (a] + x3 + (k …

Question

Let S be the sum of the first 9 terms of the
series :

{x + k$a$} + {x2 + (k + 2) (a\]} + {x3 + (k + 4) (a\]}

+ {x4 + (k + 6) (a\]} + ....  where a $$ \ne $$ 0 and x $$ \ne $$ 1.

If S = $${{{x^{10}} - x + 45a\left( {x - 1} \right)} \over {x - 1}}$$,  then k is equal to :

Accepted Answer

S = {x + k$a$} + {x2 + (k + 2) (a\]} + {x3 + (k + 4) (a\]}

+ {x4 + (k + 6) (a\]} + ....

$$S = \left( {x + {x^2} + {x^3} + ....\,9terms} \right) + $$
   $$a\left( {k + \left( {k + 2} \right) + (k + 4) + (k + 6) + ....9terms} \right)$$

$$S = {{x\left( {{x^9} - 1} \right)} \over {\left( {x - 1} \right)}} + a\left[ {{9 \over 2}\left[ {2k + \left( {9 - 1} \right)2} \right]} \right]$$

$$S = {{{x^{10}} - x} \over {x - 1}} + 9a\left( {k + 8} \right)$$

$$S = {{{x^{10}} - x + 9a\left( {k + 8} \right)\left( {x - 1} \right)} \over {x - 1}}$$

$$S = {{{x^{10}} - x + 9\left( {k + 8} \right)a\left( {x - 1} \right)} \over {x - 1}}$$

Compare with given sum, then we get

$${9\left( {k + 8} \right) = 45}$$

$$ \Rightarrow \left( {k + 8} \right) = 5$$

$$ \Rightarrow k =  - 3$$

Solutions · 1