"$$ \begin{aligned} &\text { Let }\left\{a_{n}\right\}_{n=0}^{\infty} \text { be a sequence such that } a_{0}=a_{1}=0 \text { and } \\\\ &a_{n+2}=3 a_{n+1}-2 a_{n}+1, \forall n \geq 0 . \end{aligned} $$ Then $a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24}$ is equal to"

Question

Accepted Answer

"

Given,

${a_0} = {a_1} = 0$

and ${a_{n + 2}} = 3{a_{n + 1}} - 2{a_n} + 1$

For $n = 0,\,{a_2} = 3{a_1} - 2{a_0} + 1$

$= 3\,.\,0 - 2\,.\,0 + 1$

$= 1$

For $n = 1,\,{a_3} = 3{a_2} - 2{a_1} + 1$

$= 3\,.\,1 - 2\,.\,0 + 1$

$= 4$

For $n = 2,\,{a_4} = 3{a_3} - 2{a_2} + 1$

$= 3\,.\,4 - 2\,.\,1 + 1$

$= 11$

For $n = 3,\,{a_5} = 3{a_4} - 2{a_3} + 1$

$= 3\,.\,11 - 2\,.\,4 + 1$

$= 26$

For $n = 4,\,{a_6} = 3{a_5} - 2{a_4} + 1$

$= 3\,.\,26 - 2\,.\,11 + 1$

$= 57$

$\therefore$ ${S_n} = 1 + 4 + 11 + 26 + 57\, + \,....\, + \,{t_n}$

${S_n} = 1 + 4 + 11 + 26\, + \,....\, + \,{t_{n - 1}} + {t_n}$

$0 = 1 + 3 + 7 + 15 + 31\, + \,.....\, - {t_n}$

$\Rightarrow {t_n} = 1 + 3 + 7 + 15 + 31\, + \,....$

Now, find the sum of the series,

${t_n} = 1 + 3 + 7 + 15 + 31\, + \,.....\, + \,{x_{n - 1}} + {x_n}$ .....(1)

${t_n} =$ $1 + 3 + 7 + 15\, + \,.....\, + \,{x_{n - 1}} + {x_n}$ ......(2)

Subtracting (2) from (1), we get

-------------------------------------------------------------------------

$0 = 1 + 2 + 4 + 8 + 16\, + \,....\, + \,{x_n}$

$\Rightarrow {x_n} = 1 + 2 + 4 + 8 + 16\, + \,.....\, + \,$ n terms

$= {{1({2^n} - 1)} \over {2 - 1}}$

$= {2^n} - 1$

$\therefore$ ${t_n} = \sum\limits_{n = 1}^n {{x_n}}$

$= \sum\limits_{n = 1}^n {({2^n} - 1)}$

$= \sum\limits_{n = 1}^n {{2^n} - \sum\limits_{n = 1}^n 1 }$

$= {{2({2^n} - 1)} \over {2 - 1}} - n$

$= {2^{n + 1}} - 2 - n$

${t_1} = {2^2} - 2 - 1 = 1 = {a_2}$

${t_2} = {2^3} - 2 - 2 = 4 = {a_3}$

${t_3} = {2^4} - 2 - 3 = 11 = {a_4}$

$\therefore$ ${a_{22}} = {t_{21}} = {2^{22}} - 2 - 21 = {2^{22}} - 23$

${a_{23}} = {t_{22}} = {2^{23}} - 2 - 22 = {2^{23}} - 24$

${a_{24}} = {t_{23}} = {2^{24}} - 2 - 23 = {2^{24}} - 25$

${a_{25}} = {t_{24}} = {2^{25}} - 2 - 24 = {2^{25}} - 26$

Now,

${a_{25}}{a_{23}} - 2{a_{25}}{a_{22}} - 2{a_{23}}{a_{24}} + 4{a_{22}} {a_{24}}$

$= {a_{25}}({a_{23}} - 2{a_{22}}) - 2{a_{24}}({a_{23}} - 2{a_{22}})$

$= ({a_{23}} - 2{a_{22}})({a_{25}} - 2{a_{24}})$

$= [({2^{23}} - 24) - 2({2^{22}} - 23)][({2^{25}} - 26) - 2({2^{24}} - 25)]$

$= [({2^{23}} - 24 - {2^{23}} + 46)][({2^{25}} - 26 - {2^{25}} + 50)]$

$= (22)(24)$

$= 528$

"

$$ \begin{aligned} &\text { Let }\left\{a_{n}\right\}_{n=0}^{\infty} \text { be a sequence such that } a_{0}=a_{1}=0 \text { and } \\\\ &a_{n+2}=3 a_{n+1}-2 a_{n}+1, \forall n \geq 0 . \end{aligned} $$

Then $a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24}$ is equal to

Solution

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$$ \begin{aligned} &\text { Let }\left\{a_{n}\right\}_{n=0}^{\infty} \text { be a sequence such that } a_{0}=a_{1}=0 \text { and } \\\\ &a_{n+2}=3 a_{n+1}-2 a_{n}+1, \forall n \geq 0 . \end{aligned} $$ Then $a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24}$ is equal to

Solution

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More Sequences and Series questions

Drill 25 more like these. Every day. Free.

$$ \begin{aligned} &\text { Let }\left\{a_{n}\right\}_{n=0}^{\infty} \text { be a sequence such that } a_{0}=a_{1}=0 \text { and } \\\\ &a_{n+2}=3 a_{n+1}-2 a_{n}+1, \forall n \geq 0 . \end{aligned} $$

Then $a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24}$ is equal to