Verified Solution • 2025–26 Session moderate

QJEEAdv19Ma47c9: JEE Advanced 2019 Mathematics - Class 11 JEEAdv19Ma47c9 Solution: Let and be the roots of , with > . For all positive integers n, …

Let α\alpha and β\beta be the roots of x2−x−1=0{x^2} - x - 1 = 0 , with α\alpha > β\beta . For all positive integers n, define

an=αn−βnα−β, n≥1{a_n} = {{{\alpha ^n} - {\beta ^n}} \over {\alpha - \beta }},\,n \ge 1

b1=1 and bn=an−1+an+1, n≥2{b_1} = 1\,and\,{b_n} = {a_{n - 1}} + {a_{n + 1}},\,n \ge 2

Then which of the following options is/are correct?

Multiple Choice Options

  • A ∑n=1∞bn10n=889\sum\limits_{n = 1}^\infty {{{{b_n}} \over {{{10}^n}}}} = {8 \over {89}}
  • B bn = α\alpha n + β\beta n for all n ≥\ge 1
  • C a1 + a2 + a3 + ... + an = an+2 −- 1 for all n ≥\ge 1
  • D ∑n=1∞an10n=1089\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{{10}^n}}}} = {10 \over {89}}

Step-by-Step Verified Solution

Given quadratic equation

x2−x−1=0{x^2} - x - 1 = 0 having roots α\alpha and β\beta , ( α\alpha > β\beta )

So, α=1+52\alpha = {{1 + \sqrt 5 } \over 2} and β=1−52\beta = {{1 - \sqrt 5 } \over 2}

and α+β=1\alpha + \beta = 1 , α\alpha β\beta = −1- 1

∵\because an=αn−βnα−β, n≥1{a_n} = {{{\alpha ^n} - {\beta ^n}} \over {\alpha - \beta }},\,n \ge 1

So, an+1=αn+1−βn+1α−β{a_{n + 1}} = {{{\alpha ^{n + 1}} - {\beta ^{n + 1}}} \over {\alpha - \beta }}

αn+αn−1β+αn−2β2+...+αβn−1+βn{\alpha ^n} + {\alpha ^{n - 1}}\beta + {\alpha ^{n - 2}}{\beta ^2} + ... + \alpha {\beta ^{n - 1}} + {\beta ^n}

αn−αn−2−αn−3β−...−βn−2+βn{\alpha ^n} - {\alpha ^{n - 2}} - {\alpha ^{n - 3}}\beta - ... - {\beta ^{n - 2}} + {\beta ^n}

[as α\alpha β\beta = −- 1]

= αn+βn−(αn−2+αn−3β+...+βn−2){\alpha ^n} + {\beta ^n} - ({\alpha ^{n - 2}} + {\alpha ^{n - 3}}\beta + ... + {\beta ^{n - 2}})

= αn+βn−an−1{\alpha ^n} + {\beta ^n} - {a_{n - 1}}

[as an−1=αn−1−βn−1α−β=αn−2+αn−3β+...+βn−2]\left[ {as\,{a_{n - 1}} = {{{\alpha ^{n - 1}} - {\beta ^{n - 1}}} \over {\alpha - \beta }} = {\alpha ^{n - 2}} + {\alpha ^{n - 3}}\beta + ... + {\beta ^{n - 2}}} \right]

⇒\Rightarrow an+1+an−1=αn+βn=bn, ∀n≥1{a_{n + 1}} + {a_{n - 1}} = {\alpha ^n} + {\beta ^n} = {b_n},\,\forall n \ge 1

So, option (b) is correct.

Now, ∑n=1∞bn10n=∑n=1∞αn+βn10n\sum\limits_{n = 1}^\infty {{{{b_n}} \over {{{10}^n}}}} = \sum\limits_{n = 1}^\infty {{{{\alpha ^n} + {\beta ^n}} \over {{{10}^n}}}}

[as, bn = α\alpha n + β\beta n]

= ∑n=1∞(α10)n+∑n=1∞(β10)n\sum\limits_{n = 1}^\infty {{{\left( {{\alpha \over {10}}} \right)}^n}} + \sum\limits_{n = 1}^\infty {{{\left( {{\beta \over {10}}} \right)}^n}}

∵\because \left[ {\left| {{\alpha \over {10}}} \right| < 1\,and\,\left| {{\beta \over {10}}} \right| < 1} \right]

=α101−α10+β101−β10=α10−α+β10−β= {{{\alpha \over {10}}} \over {1 - {\alpha \over {10}}}} + {{{\beta \over {10}}} \over {1 - {\beta \over {10}}}} = {\alpha \over {10 - \alpha }} + {\beta \over {10 - \beta }}

=10α−αβ+10β−αβ(10−α)(10−β)= {{10\alpha - \alpha \beta + 10\beta - \alpha \beta } \over {(10 - \alpha )(10 - \beta )}}

=10(α+β)−2αβ100−10(α+β)+αβ= {{10(\alpha + \beta ) - 2\alpha \beta } \over {100 - 10(\alpha + \beta ) + \alpha \beta }}

=10(1)−2(−1)100−10(1)−1= {{10(1) - 2( - 1)} \over {100 - 10(1) - 1}}

[as α\alpha + β\beta = 1 and α\alpha β\beta = −- 1]

=1289= {{12} \over {89}}

So, option (a) is not correct.

∵\because α2=α+1{\alpha ^2} = \alpha + 1 and β2=β+1{\beta ^2} = \beta + 1

⇒\Rightarrow αn+2=αn+1+αn{\alpha ^{n + 2}} = {\alpha ^{n + 1}} + {\alpha ^n} and βn+2=βn+1+βn{\beta ^{n + 2}} = {\beta ^{n + 1}} + {\beta ^n}

⇒\Rightarrow (αn+2+βn+2)=(αn+1+βn+1)+(αn+βn)({\alpha ^{n + 2}} + {\beta ^{n + 2}}) = ({\alpha ^{n + 1}} + {\beta ^{n + 1}}) + ({\alpha ^n} + {\beta ^n})

⇒an+2=an+1+an\Rightarrow {a_{n + 2}} = {a_{n + 1}} + {a_n}

Similarly, an+1=an+an−1{a_{n + 1}} = {a_n} + {a_{n - 1}}

an=an−1+an−2{a_n} = {a_{n - 1}} + {a_{n - 2}}

..............

............

a3=a2+a1{a_3} = {a_2} + {a_1}

On adding, we get

an+2=(an+an−1+an−2+...+a2+a1)+a2{a_{n + 2}} = ({a_n} + {a_{n - 1}} + {a_{n - 2}} + ... + {a_2} + {a_1}) + {a_2}

∵\because [a2=α2−β2α−β=α+β=1]\left[ {{a_2} = {{{\alpha ^2} - {\beta ^2}} \over {\alpha - \beta }} = \alpha + \beta = 1} \right]

So, an+2−1=a1+a2+a3+......+an{a_{n + 2}} - 1 = {a_1} + {a_2} + {a_3} + ...... + {a_n}

So, option (c) is also correct.

And, now ∑n=1∞an10n=∑n=1∞αn−βn(α−β)10n\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{{10}^n}}}} = \sum\limits_{n = 1}^\infty {{{{\alpha ^n} - {\beta ^n}} \over {(\alpha - \beta ){{10}^n}}}}

=1α−β[∑n=1∞(a10)n−∑n=1∞(β10)n]= {1 \over {\alpha - \beta }}\left[ {\sum\limits_{n = 1}^\infty {{{\left( {{a \over {10}}} \right)}^n}} - \sum\limits_{n = 1}^\infty {{{\left( {{\beta \over {10}}} \right)}^n}} } \right]

=1α−β[α101−α10−β101−β10]= {1 \over {\alpha - \beta }}\left[ {{{{\alpha \over {10}}} \over {1 - {\alpha \over {10}}}} - {{{\beta \over {10}}} \over {1 - {\beta \over {10}}}}} \right] ,

\left[ {as\left| {{\alpha \over {10}}} \right| < 1\,and\,\left| {{\beta \over {10}}} \right| < 1} \right]

=10(α−β)(α−β)[100−10(α+β)+αβ]= {{10(\alpha - \beta )} \over {(\alpha - \beta )[100 - 10(\alpha + \beta ) + \alpha \beta ]}}

=10100−10−1=1089= {{10} \over {100 - 10 - 1}} = {{10} \over {89}}

Hence, options (b), (c) and (d) are correct.

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