Sequences and Series

Taylor series expansion of $e^{-x}$ is

$ e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\ldots=\sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k!} $

Now, $\frac{1-a x-x^2}{e^x}=\left(1-a x-x^2\right) e^{-x}$

                           $ =e^{-x}-(a x) e^{-x}-\left(x^2\right) e^{-x} $

From $e^{-x}$, the coefficient of $x^n$ in $e^{-x}$ is $\frac{(-1)^n}{n!}$

From $-a x e^{-x}$, the coefficient of $x^n$ in $-a x e^{-x}$ is given by the coefficient of $x^{n-1}$ in $e^{-x}$, multiplied by $-a$

$ -a \cdot \frac{(-1)^{n-1}}{(n-1)!} $

From $-x^2 e^{-x}$, the coefficient of $x^n$ in $-x^2 e^{-x}$ is given by the coefficient of $x^{n-2}$ in $e^{-x}$, multiplied by -1 :

$ \frac{-(-1)^{n-2}}{(n-2)!} $

Now, combine the coefficient from all three terms

$ \begin{aligned} & \text { coefficient of } x^n=\frac{(-1)^n}{n!}-a \frac{(-1)^{n-1}}{(n-1)!}-\frac{(-1)^{n-2}}{(n-2)} \\ & =(-1)^{n-2}\left[\frac{(-1)^2}{n!}-a \cdot \frac{(-1)}{(n-1)!}-1 \cdot \frac{1}{(n-2)!}\right] \\ & =(-1)^{n-2}\left[\frac{1}{n!}+\frac{a}{(n-1)!}-\frac{1}{(n-2)!}\right] \\ & =\frac{(-1)^n}{n!}\left[1+(a+1) n-n^2\right] \end{aligned} $

Given, $a+b+c=25\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, $2, a, b$ are in AP

$ \Rightarrow \quad 2 a=2+b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

And $b, c, 18$ are in GP

$ \Rightarrow \quad c^2=18 b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

From Eqs. (i) and (ii), we get

$ \begin{aligned} & \frac{2+b}{2}+b+c=25 \\ & \Rightarrow 2+b+2 b+2 c=50 \\ & \Rightarrow 3 b+2 c=48 \\ & \Rightarrow 3\left(\frac{c^2}{18}\right)+2 c=48\,\,\,\,\,[using\,\, Eq. (iii)] \\ \Rightarrow & c^2+12 c-288=0 \Rightarrow c=12-2 \end{aligned} $

$\because a, b, c$ lies between 2 and 18

$ \therefore \quad c=12 $

Thus, $a=5$ and $b=8$

$ \therefore \quad a+b+c=5+8+12=25 $

Using $A M>G M$, we have

$ \begin{aligned} & \frac{a+b+c}{3}>(a b c)^{1 / 3} \\ \Rightarrow & a+b+c>3(a b c)^{1 / 3} \Rightarrow(s-d)>3(a b c)^{1 / 3} \end{aligned} $

Similarly, $(s-a)>3(b c d)^{1 / 3},(s-b)>3(a c d)^{1 / 3}$ and

$ \begin{aligned} & (s-c)>3(a b d)^{1 / 3} \\ & \therefore(s-a)(s-b)(s-c)(s-d)> \\ & \quad 3^4(a b c)^{\frac{1}{3}}(b c d)^{\frac{1}{3}}(a c d)^{\frac{1}{3}}(a b d)^{\frac{1}{3}} \\ & \quad=81 a b c d \end{aligned} $

Thus, $k=81$

We know that $ \mathrm{AM} \geq \mathrm{GM} $

Now, applying Eq. (i) in $ a $ and $ b $

$ \Rightarrow \quad \frac{a+b}{2} \geq \sqrt{a b} $

applying Eq. (i) in $ b $ and $ c $

$ \Rightarrow \quad \frac{b+c}{2} \geq \sqrt{b c} $

and applying Eq. (i) in $ c $ and $ a $

$ \Rightarrow \quad \frac{c+a}{2} \geq \sqrt{c a} $

By Eqs. (ii), (iii) and'(iv)

$ \Rightarrow \quad \frac{(a+\bar{b})(b+c)(c+a)}{8} \geq a b c $

$ \Rightarrow \quad(a+b)(b+c)(c+a) \geq 8 a b c $

Let $ S=\sum\limits_{k=1}^{n} k(k+1)(k-1) $

$ =\sum\limits_{k=1}^{n} k^{3}-k $

We know that

$ \sum\limits_{r=1}^{P} r=\frac{P(P+1)}{2} $

and $ \quad \sum\limits_{r=1}^{P} r^{3}=\left[\frac{P(P+1)}{2}\right]^{2} $

$ \begin{aligned} \therefore & =\left[\frac{n(n+1)}{2}\right]^{2}-\left[\frac{n(n+1)}{2}\right] \\ & =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}-1\right] \\ & =\left(\frac{n^{2}+n}{2}\right)\left[\frac{n^{2}+n-2}{2}\right] \\ & =\frac{n^{4}}{4}+\frac{n^{3}}{2}-\frac{n^{2}}{4}-\frac{n}{2} \end{aligned} $

Now, by comparing $ p n^{4}+q n^{3}+t n^{2}+s n $ with Eq. (i),

$ S=-\frac{1}{2} $

To solve the problem, let's consider the setup described:

The three numbers in an Arithmetic Progression (AP) are denoted as $a$, $a+6$, and $a+12$, where the common difference is 6.

The sequence of four numbers has the first and last numbers equal.

Thus, the four numbers are $(a+12)$, $a$, $a+6$, and $(a+12)$.

The problem states that the first three numbers are in a Geometric Progression (GP). Therefore, the following relationship should hold:

$ a^2 = (a+12)(a+6) $

Expanding and solving the equation:

$ a^2 = a^2 + 18a + 72 $

Simplifying:

$ 0 = 18a + 72 $

Solving for $a$:

$ 18a = -72 $

$ a = -\frac{72}{18} = -4 $

Given this value of $a$, the four numbers are calculated as:

$a+12 = 8$,

$a = -4$,

$a+6 = 2$,

$a+12 = 8$.

Therefore, the sequence of four numbers is $8, -4, 2, 8$.

The two intermediate numbers among the four (excluding the repeated elements at the start and end) are $-4$ and $2$. Hence, the two other numbers in this sequence are $-4, 2$.

Given, $ \frac{e^{7 x}+e^{x}}{e^{3 x}} \Rightarrow e^{4 x}+e^{-2 x} $

Series of $ e^{a}=1+\frac{a}{1!}+\frac{a^{2}}{2!}+\frac{a^{3}}{3!} \ldots $

$ \begin{array}{l} \Rightarrow e^{4 x}+e^{-2 x}=\left(1+\frac{4 x}{1!}+\frac{(4 x)^{2}}{2!}+\frac{(4 x)^{3}}{3!}+\ldots\right) +\left(1+\frac{(-2 x)}{1!}+\frac{(-2 x)^{2}}{2!}+\ldots \ldots\right) \end{array} $

Here, coefficient of

$ \begin{array}{c} x^{2} \equiv \frac{4^{2}}{2!}+\frac{(-2)^{2}}{2!} \\ x^{3} \equiv \frac{4^{3}}{3!}+\frac{(-2)^{3}}{3!} \\ \vdots \\ x^{n}=\frac{4^{n}}{n!}+\frac{(-2)^{n}}{n!} \end{array} $

To solve the problem, let’s analyze it step by step.

Firstly, since $\frac{1}{16}, a,$ and $b$ are in geometric progression (GP), the common ratio can be denoted as $r$. For GP, the relation between terms is given by:

If $\frac{1}{16}$ is the first term and $a$ is the second term, then:

$ a = \frac{1}{16} \cdot r $

If $a$ is the second term and $b$ is the third term, then:

$ b = a \cdot r $

Substituting the expression for $a$ in the second equation:

$ b = \left( \frac{1}{16} \cdot r \right) \cdot r = \frac{1}{16} \cdot r^2 $

We are also given that $\frac{1}{a}, \frac{1}{b},$ and $6$ are in arithmetic progression (AP). For AP, the middle term is the average of the two surrounding terms:

Since $\frac{1}{a}, \frac{1}{b}, 6$ are in AP:

$ 2 \cdot \frac{1}{b} = \frac{1}{a} + 6 $

Now, substituting the values of $\frac{1}{a}$ and $\frac{1}{b}$ from the GP relations:

Since $ a = \frac{1}{16} \cdot r $:

$ \frac{1}{a} = \frac{1}{ \left( \frac{1}{16} \cdot r \right) } = 16 \cdot \frac{1}{r} = \frac{16}{r} $

Also, since $ b = \frac{1}{16} \cdot r^2 $:

$ \frac{1}{b} = \frac{1}{ \left( \frac{1}{16} \cdot r^2 \right) } = 16 \cdot \frac{1}{r^2} = \frac{16}{r^2} $

Substitute these values into the AP equation:

$ 2 \cdot \frac{16}{r^2} = \frac{16}{r} + 6 $

Expanding and simplifying, we get:

$ \frac{32}{r^2} = \frac{16}{r} + 6 $

To clear the fractions, multiply all terms by $r^2$:

$ 32 = 16r + 6r^2 $

Rearranging into standard quadratic form:

$ 6r^2 + 16r - 32 = 0 $

Divide throughout by 2 for simplicity:

$ 3r^2 + 8r - 16 = 0 $

Solving the quadratic equation using the quadratic formula:

$ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

where $a = 3$, $b = 8$, and $c = -16$:

$ r = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot (-16)}}{2 \cdot 3} = \frac{-8 \pm \sqrt{64 + 192}}{6} = \frac{-8 \pm \sqrt{256}}{6} = \frac{-8 \pm 16}{6} $

This gives us two possible values for $r$:

$ r = \frac{8}{6} = \frac{4}{3} $

$ r = \frac{-24}{6} = -4 $

Since $a$ and $b$ are positive, $r = \frac{4}{3}$ is the suitable solution.

Now calculate values for $a$ and $b$:

$ a = \frac{1}{16} \cdot \frac{4}{3} = \frac{1}{12} $

$ b = \left( \frac{1}{12} \right) \cdot \frac{4}{3} = \frac{1}{9} $

Finally,$ 72(a + b) $ =:

$ 72 \left( \frac{1}{12} + \frac{1}{9} \right) = 72 \left( \frac{3 + 4}{36} \right) = 72 \left( \frac{7}{36} \right) = 14 $

Therefore, the answer is Option B: 14.

Given, $a_1, a_2, \ldots a_n$ are in HP

$\Rightarrow \frac{1}{a_1}, \frac{1}{a_2}, \ldots, \frac{1}{a_n}$ are in AP

$\Rightarrow \frac{1}{a_2}-\frac{1}{a_1}=d$, where $d$ is common difference

$\begin{aligned} \Rightarrow \quad a_1 a_2 & =\frac{a_1-a_2}{d}=\frac{a_1}{d}-\frac{a_2}{d} \quad \text{... (i)}\\ & a_2 a_3=\frac{a_2}{d}-\frac{a_3}{d} \quad \text{... (ii)} \end{aligned}$

$a_{n-1} a_n=\frac{a_{n-1}}{d}-\frac{a_n}{d} \ldots(n-1)$

On adding Eqs. (i), (ii), ......, $(n-1)$, we get

$\begin{array}{r} a_1 a_2+a_2 a_3+\ldots+a_{n-1} a_n \\ =\frac{a_1}{d}-\frac{a_n}{d} \end{array}$

$\begin{aligned} & \text { Also, } \frac{1}{a_n}=\frac{1}{a_1}+(n-1) d \\ & \Rightarrow \quad \frac{a_1-a_n}{d}=(n-1) a_1 a_n \\ & \therefore \quad a_1 a_2+a_2 a_3+\ldots+a_{n-1} a_n=(n-1) a_1 a_n \end{aligned}$

Let $a, b, c$ and $d$ be four numbers of sequence.

Now, according to the question, $b^2=a c$ and $c-b=d-c=6$

Also given that $a=d$

$\therefore b^2=a c \Rightarrow b^2=a\left(\frac{a+b}{2}\right) \quad\left[\because c=\frac{b+d}{2} \text { and } a=d\right]$

$\Rightarrow a^2-2 b^2+a b=0$

Now, $c-b=6$ and $a-c=6$, gives $a-b=12$

$\begin{array}{ll} \Rightarrow & b=a-12 \\ \therefore & a^2-2 b^2+a b=0 \\ \Rightarrow & a^2-2(a-12)^2+a(a-12)=0 \\ \Rightarrow & a^2-2 a^2-288+48 a+a^2-12 a=0 \\ \Rightarrow & 36 a=288 \Rightarrow a=8 \end{array}$

Hence, last term is $d=a=8$

Let d be the common difference of the AP. Then,

${a_{10}} = 3 \Rightarrow {a_1} + 9d = 3$

$ \Rightarrow 2 + 9d = 3 \Rightarrow d = {1 \over 9}$

$\therefore$ ${a_4} = {a_1} + 3d = 2 + {1 \over 3} = {7 \over 3}$

Let D be the common difference of ${1 \over {{h_1}}},{1 \over {{h_2}}},\,...\, - {1 \over {{h_{10}}}}$.

Then, ${h_{10}} = 3$

$ \Rightarrow {1 \over {{h_{10}}}} = {1 \over 3} \Rightarrow {1 \over 2} + 9D = {1 \over 3}$

$ \Rightarrow 9D = - {1 \over 6} \Rightarrow D = - {1 \over {54}}$

$\therefore$ ${1 \over {{h_7}}} = {1 \over {{h_1}}} + 6D = {1 \over 2} - {1 \over 9} = {7 \over {18}}$

$ \Rightarrow {h_7} = {{18} \over 7}$

$\therefore$ ${a_4}{h_7} = {7 \over 3} \times {{18} \over 7} = 6$

It is given that a₁, a₂, a₃ .... are in HP.

Therefore, ${1 \over {{a_1}}},{1 \over {{a_2}}},{1 \over {{a_3}}}$, ..... are in AP

Let d be the common difference of the AP.

$\therefore$ ${1 \over {{a_n}}} = {1 \over {{a_1}}} + (n - 1)d$ and ${1 \over {{a_{20}}}} = {1 \over {{a_1}}} + 19d$

$ \Rightarrow {1 \over {{a_n}}} = {1 \over 5} + (n - 1)d$ and ${1 \over {25}} = {1 \over 5} + 19d$

$ \Rightarrow {1 \over {{a_n}}} = {1 \over 5} + (n - 1)d$ and $d = {{ - 4} \over {19 \times 25}}$

$ \Rightarrow {1 \over {{a_n}}} = {1 \over 5} - {{4(n - 1)} \over {19 \times 25}}$

$ \Rightarrow {1 \over {{a_n}}} = {{95 - 4n + 4} \over {19 \times 25}}$

$ \Rightarrow {a_n} = {{99 - 4n} \over {19 \times 25}}$

Now, ${a_n} < 0$

$ \Rightarrow {{99 - 4n} \over {19 \times 25}} < 0$

$ \Rightarrow 99 - 4n < 0$

$ \Rightarrow 4n > 99$

$ \Rightarrow n > 24{3 \over 4} \Rightarrow n \ge 25$

Since, the first 11 terms are in AP, d = 2

$\therefore$ a₁₁ = a + 10d = a + 20

the middle term of AP is

T₆ = a + 5d = a + 10

For the next 11 terms in GP

r = 2

Given, a + 2b + 3c = 12 $\forall$ a, b, c $\in$ R⁺

As, AM $\ge$ GM

${{a + b + b + c + c + c} \over 6} \ge \root 6 \of {a{b^2}{c^3}} $

$ \Rightarrow {{12} \over 6} \ge \root 6 \of {a{b^2}{c^3}} \Rightarrow a{b^2}{c^3} \le {2^6}$

Hence, the maximum value of ab²c³ is 2⁶.

a_n of the series $ = n{(2n + 1)^2}\forall n \in E$

${a_n} = n[4{n^2} + 4n + 1] \Rightarrow {a_n} = 4{n^3} + 4{n^2} + n$

${S_n} = \sum {{a_n} = 4\sum {{n^3} + 4\sum {{n^2} + \sum n } } } $

$ = 4{\left( {{{n(n + 1)} \over 2}} \right)^2} + {{4n(n + 1)(2n + 1)} \over 6} + {{n(n + 1)} \over 2}$

$ = {{n(n + 1)} \over 2}\left[ {2{n^2} + 2n + {{4(2n + 1)} \over 3} + 1} \right]$

$ = {{n(n + 1)} \over 2}\left[ {{{6{n^2} + 6n + (8n + 4) + 3} \over 3}} \right]$

$ = {{n(n + 1)} \over 2}\left[ {{{6{n^2} + 14n + 7} \over 3}} \right] = {n \over 6}(n + 1)(6{n^2} + 14n + 7)$

$\because$ 13, a₁, a₂, ....., a₂₀, 67 are in AP.

$\therefore$ a₁ + a₂ + a₃ + ..... + a₂₀ = 20$\left( {{{13 + 67} \over 2}} \right)$ = 800

Also, AM > GM

$\Rightarrow$ ${{{a_1} + {a_2} + .... + {a_{20}}} \over {20}} \ge {({a_1},{a_2},{a_3},....,{a_{20}})^{{1 \over {20}}}}$

$ \Rightarrow 40 \ge {({a_1}\,.\,{a_2}\,.\,{a_3}\,....\,{a_{20}})^{{1 \over {20}}}}$

Hence, maximum value of ${a_1}\,.\,{a_2}\,.\,{a_3}\,....\,{a_{20}}$ is (40)²⁰.

Since, p, q and r are positive and are in AP.

$\therefore$ $q = {{p + r} \over 2}$ ...... (i)

$\because$ The roots of the equation, px² + qx + r = 0 are real.

$ \Rightarrow {q^2} \ge 4pr \Rightarrow {\left[ {{{p + r} \over 2}} \right]^2} \ge 4pr$ [from Eq. (i)]

$ \Rightarrow {p^2} + {r^2} - 14pr \ge 0$

$ \Rightarrow {\left( {{r \over p}} \right)^2} - 14\left( {{r \over p}} \right) + 1 \ge 0$

$ \Rightarrow {\left( {{r \over p} - 7} \right)^2} - 48 \ge 0$

$ \Rightarrow {\left( {{r \over p} - 7} \right)^2} - {(4\sqrt 3 )^2} \ge 0 \Rightarrow \left| {{r \over p} - 7} \right| \ge 4\sqrt 3 $

Since, $g = \sqrt {ab} $. Also, a, p, q and b are in AP.

So, common difference d is ${{b - a} \over 3}$.

$\therefore$ $p = a + d = a + {{b - a} \over 3} = {{2a + b} \over 3}$

$q = b - d = b - {{b - a} \over 3} = {{a + 2b} \over 3}$

Now, $(2p - q)(p - 2q)$

$ = {{(4a + 2b - a - 2b)} \over 3}\,.\,{{(2a + b - 2a - 4b)} \over 3}$

$ = - ab = - {g^2}$

Since x, y, z are three consecutive integers

$\therefore$ 2y = x + z

$\Rightarrow$ (4y²) = (x + z)²

= (x $-$ z)² + 4xz

= ($-$2)² + 4xz [$\because$ x $-$ z = $-$2]

$\Rightarrow$ 4y² = 4 + 4xz

$\Rightarrow$ y² = 1 + xz

Now,

${1 \over 2}{\log _e}x + {1 \over 2}{\log _e}z + {1 \over {2xz + 1}} + {1 \over 3}{\left( {{1 \over {2xz + 1}}} \right)^3} + ...$

$ = {1 \over 2}\left[ {{{\log }_e}(xz) + 2\left( {{1 \over {2zx + 1}} + {1 \over 3}\,.\,{1 \over {{{(2zx + 1)}^3}}} + ...} \right)} \right]$

$ = {1 \over 2}\left[ {{{\log }_e}(xz) + {{\log }_e}\left( {{{1 + {1 \over {2zx + 1}}} \over {1 - {1 \over {2zx + 1}}}}} \right)} \right]$

$ = {1 \over 2}\left[ {{{\log }_e}(xz) + {{\log }_e}\left( {{{1 + xz} \over {xz}}} \right)} \right]$

$ = {1 \over 2}[{\log _e}(1 + xz)] = {1 \over 2}[{\log _e}{(y)^2}] = {\log _e}(y)$

We have, ${S_\infty } = \sum\limits_{k = 1}^\infty {\sum\limits_{n = 1}^\infty {{k \over {{2^{n + k}}}}} } $

${S_\infty } = \sum\limits_{k = 1}^\infty {{k \over {{2^k}}}\,.\,\sum\limits_{n = 1}^\infty {{1 \over {{2^n}}}} } $

${S_\infty } = \sum\limits_{k = 1}^\infty {{k \over {{2^k}}}\left( {{1 \over 2} + {1 \over {{2^2}}} + {1 \over {{2^3}}} + ....} \right)} $

${S_\infty } = \sum\limits_{k = 1}^\infty {{k \over {{2^k}}}\left[ {{1 \over 2}\left( {1 + {1 \over 2} + {1 \over {{2^2}}}...} \right)} \right]} $

${S_\infty } = \sum\limits_{k = 1}^\infty {{k \over {{2^k}}}\left( {{1 \over 2} \times {1 \over {1 - {1 \over 2}}}} \right) = \sum\limits_{k = 1}^\infty {{k \over {{2^k}}}} } $

${S_\infty } = {1 \over 2} + {2 \over {{2^2}}} + {3 \over {{2^3}}} + {4 \over {{2^4}}} + ...$ ..... (i)

${1 \over 2}{S_\infty } = {1 \over {{2^2}}} + {2 \over {{2^3}}} + {3 \over {{2^4}}} + ...$ ...... (ii)

From Eqs. (i) and (ii), we get

${1 \over 2}{S_\infty } = {1 \over 2} + {1 \over {{2^2}}} + {1 \over {{2^3}}} + ...$

${1 \over 2}{S_\infty } = {1 \over 2}\left( {{1 \over {1 - {1 \over 2}}}} \right)$

${1 \over 2}{S_\infty } = 1$

${S_\infty } = 2$