The sum $1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\ldots$ upto $\infty$ terms, is equal to

The series given is: $ 1 + \frac{1+3}{2!} + \frac{1+3+5}{3!} + \frac{1+3+5+7}{4!} + \ldots $ In general, the $ r $-th term of the series, denoted as $ T_r $, takes the form: $ T_r = \frac{r^2}{r!} $ To simplify $ T_r $, we write it in terms of factorials: $ T_r = \frac{r}{(r-1)!} = \frac{(r-1)+1}{(r-1)!} = \frac{1}{(r-2)!} + \frac{1}{(r-1)!} $ Thus, the series can be expressed as: $ \sum\limits_{r=1}^{\infty} T_r = \sum\limits_{r=1}^{\infty} \left( \frac{1}{(r-2)!} + \frac{1}{(r-1)!} \right) $ T

Sequences and Series — Page 9 | JEE Mathematics

Q81

The sum

1+3+11+25+45+71+\ldots

upto 20 terms, is equal to

A 7240

B 8124

C 7130

D 6982

Correct Answer

Option A

Solution

\begin{aligned} &T_r=3 r^2-7 r+5 \text{ using second order difference }\\ &\begin{aligned} & \sum_{r=1}^{20}\left(3 r^2-7 r+5\right)=3 \Sigma r^2-7 \Sigma r+5 \Sigma(1) \\ & =\frac{3(n)(n+1)(2 n+1)}{6}-\frac{7 n(n+1)}{2}-5 n, n=20 \\ & =7240 \end{aligned} \end{aligned}

Q82

Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and

q

respectively. Let d and D be the common differences of

\mathrm{AP}^{\prime} \mathrm{s}

in

A

and

B

respectively such that

D=d+3, d>0

. If

\dfrac{p+q}{p-q}=\dfrac{19}{5}

, then

\mathrm{p}-\mathrm{q}

is equal to

A 540

B 450

C 600

D 630

Correct Answer

Option A

Solution

Let the terms in $A$ be $a_1-d, a_1, a_1+d$ and in $B$ be $a_2-D, a_2, a_2+D$ Now $3 a_1=36$

\Rightarrow \quad a_1=12

and $3 a_2=36$

\Rightarrow \quad a_2=12

Now $(12-d)(12)(12+d)=p$ and $(12-D)(12)(12+D)=q$ Also $\dfrac{p+q}{p-q}=\dfrac{19}{5}$

\begin{aligned} & \Rightarrow 12 q=7 p \\ & \Rightarrow 12(12-D)(12)(12+D)=7(12-d)(12)(12+\mathrm{d}) \\ & \Rightarrow 12(9-d)(12)(15-d)=7(12-d)(12)(12+\mathrm{d}) \\ & \Rightarrow 12\left(135-d^2-6 d\right)=7\left(144-d^2\right) \\ & \Rightarrow d=6, D=9 \\ & p=6 \times 12 \times 18=1296 \\ & q=756 \\ & p-q=540 \end{aligned}

Q83

Let

A=\{1,6,11,16, \ldots\}

and

B=\{9,16,23,30, \ldots\}

be the sets consisting of the first 2025 terms of two arithmetic progressions. Then

n(A \cup B)

is

A 3814

B 4003

C 4027

D 3761

Correct Answer

Option D

Solution

\begin{array}{ll} 1^{\text{st }} \text{ A.P. }: 1,6,11 \ldots & \Rightarrow T_n=S_n-4 \\ 2^{\text{nd }} \text{ A.P.: } 9,16,23 \ldots & \Rightarrow T_m=2+7 m \end{array}

Let's find when they are equal for the first time:

\begin{aligned} & 5 n-4=2+7 m \\ & \Rightarrow 5 n-7 m=6 \\ & \Rightarrow n=4, m=2 \end{aligned}

$\Rightarrow 16$ is the first term, common difference will be

\operatorname{LCM}\left(d_1, d_2\right)=\operatorname{LCM}(5,7)=35

$\Rightarrow$ Common terms will be 16, 51, $86 \ldots$ The last term of $1^{\text{st }}$ A.P.

=T_{2025}=5 \times 2025-4=10121

\begin{aligned} &\Rightarrow \text{ Common term must be less than that }\\ &\begin{aligned} & \Rightarrow 35 n-19 \\ & \Rightarrow 35 n-19 \leq 10121 \Rightarrow 35 n \leq 10140 \\ & \quad \Rightarrow n \leq 289.7 \\ & \quad \Rightarrow n=289 \\ & \Rightarrow \operatorname{in} n(A \cup B)=n(A)+n(B)-n(A \cap B) \\ & \quad=2025+2025-289 \\ & \quad=3761 \end{aligned} \end{aligned}

Q84

1+3+5^2+7+9^2+\ldots

upto 40 terms is equal to

A 40870

B 41880

C 43890

D 33980

Correct Answer

Option B

Solution

Step 1: Breaking the sequence into two parts The given sequence is $1+3+5^2+7+9^2+\ldots$ up to 40 terms.

Let's group the terms as follows: The terms are arranged like this: $1$ , $3$ , $5^2$ , $7$ , $9^2$ , $11$ , $13^2$ , $15$ , $17^2$ , ...

You can see that every third term is being squared, while others are just numbers.

So, split the sequence into: all squared terms ( $1^2, 5^2, 9^2, ...$ ) and all other numbers ( $3, 7, 11, ...$ ).

Step 2: Counting the terms in each group Out of 40 terms, half will be squared terms and half will be non-squared terms.

So we have 20 squared terms and 20 ordinary numbers.

Step 3: Sums for each group The squared numbers follow this pattern: $1^2, 5^2, 9^2, ...$ , which can be written as $(4k-3)^2$ for $k$ from 1 to 20.

The other numbers are $3, 7, 11, ..., (4k-1)$ for $k$ from 1 to 20.

Step 4: Write these as formulas Total sum = Sum of squares part $+$ Sum of ordinary numbers part.

The sum of the squared numbers is $\sum_{k=1}^{20} (4k-3)^2$ .

The sum of the other numbers (arithmetic series): $3 + 7 + 11 + ...$ for 20 terms.

The first term is 3, the last term is $3 + (20-1)\times4 = 3 + 76 = 79$ .

So, the sum is $\dfrac{20}{2} [3 + 79] = 10 \times 82 = 820$ .

Step 5: Expanding and simplifying Expand $(4k-3)^2$ :

(4k-3)^2 = 16k^2 - 24k + 9

So, sum up these for $k=1$ to $20$ : $16 \sum k^2 - 24 \sum k + 9 \times 20$ We know formulas: $\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$ $\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$ Plug in $n=20$ : $\sum_{k=1}^{20} k^2 = \dfrac{20 \times 21 \times 41}{6}$ $\sum_{k=1}^{20} k = \dfrac{20 \times 21}{2}$ Now, sum up: $= 16 \left(\dfrac{20 \times 21 \times 41}{6}\right) - 24\left(\dfrac{20 \times 21}{2}\right) + 180 + 820$ Step 6: Calculating $16 \left(\dfrac{20 \times 21 \times 41}{6}\right) = 16 \times 2870 = 45920$ $24 \left(\dfrac{20 \times 21}{2}\right) = 24 \times 210 = 5040$ $9 \times 20 = 180$ Sum of the other numbers = $820$ So, total = $45920 - 5040 + 180 + 820$ $= 45920 - 5040 = 40880$ $40880 + 180 = 41060$ $41060 + 820 = 41880$ Final Answer The sum of the 40 terms is $41880$ .

Q85

The sum

1+\dfrac{1+3}{2!}+\dfrac{1+3+5}{3!}+\dfrac{1+3+5+7}{4!}+\ldots

upto

\infty

terms, is equal to

A

3 e

B

2 e

C

4 e

D

6 e

Correct Answer

Option B

Solution

The series given is: $1 + \dfrac{1+3}{2!} + \dfrac{1+3+5}{3!} + \dfrac{1+3+5+7}{4!} + \ldots$ In general, the $r$ -th term of the series, denoted as $T_r$ , takes the form: $T_r = \dfrac{r^2}{r!}$ To simplify $T_r$ , we write it in terms of factorials: $T_r = \dfrac{r}{(r-1)!} = \dfrac{(r-1)+1}{(r-1)!} = \dfrac{1}{(r-2)!} + \dfrac{1}{(r-1)!}$ Thus, the series can be expressed as: $\sum\limits_{r=1}^{\infty} T_r = \sum\limits_{r=1}^{\infty} \left( \dfrac{1}{(r-2)!} + \dfrac{1}{(r-1)!} \right)$ This breaks into two parts: $= \sum\limits_{r=1}^{\infty} \dfrac{1}{(r-2)!} + \sum\limits_{r=1}^{\infty} \dfrac{1}{(r-1)!}$ Each part is a well-known series.

Specifically: $\sum\limits_{r=1}^{\infty} \dfrac{1}{(r-2)!} = e$ , starting from the term where $r-2 = 0$ , which aligns with the expansion of $e$ . $\sum\limits_{r=1}^{\infty} \dfrac{1}{(r-1)!} = e$ , for the terms starting where $r-1 = 0$ .

Consequently, the sum of the entire series is: $e + e = 2e$

Q86

The sum of series

{1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}} - .......

upto infinity is

A

{e^{ - {1 \over 2}}}

B

{e^{ + {1 \over 2}}}

C

{e^{ - 2}}

D

{e^{ - 1}}

Correct Answer

Option D

Solution

We know that

{e^x} = 1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + ........\infty

Put

x=-1

$\therefore$

{e^{ - 1}} = 1 - 1 + {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}}..........\infty

$\therefore$

{e^{ - 1}} = {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}} - {1 \over {5!}}........\infty

Q87

The first two terms of a geometric progression add up to 12. the sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is

A - 4

B - 12

C 12

D 4

Correct Answer

Option B

Solution

As per question,

\,\,\,\,\,\,\,\,\,\,\,\,a + ar = 12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

\,\,\,\,\,\,\,\,\,\,\,\,a{r^2} + a{r^3} = 48\,\,\,\,\,\,\,\,\,...\left( 2 \right)

\Rightarrow {{a{r^2}\left( {1 + r} \right)} \over {a\left( {1 + r} \right)}} = {{48} \over {12}}

\Rightarrow {r^2} = 4, \Rightarrow r = - 2

(As terms are

=+ve

and

-ve

alternately)

\Rightarrow a = - 12

Q88

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777,........,is

A

{7 \over {81}}\left( {179 - {{10}^{ - 20}}} \right)

B

\,{7 \over 9}\left( {99 - {{10}^{ - 20}}} \right)

C

{7 \over {81}}\left( {179 + {{10}^{ - 20}}} \right)

D

{7 \over 9}\left( {99 + {{10}^{ - 20}}} \right)

Correct Answer

Option C

Solution

Given sequence can be written as

{7 \over {10}} + {{77} \over {100}} + {{777} \over {{{10}^3}}} + ..... +

up to

20

terms

= 7\left[ {{1 \over {10}} + {{11} \over {100}} + {{111} \over {{{10}^3}}} + ...... + } \right.\,\,

up to

20

terms ] Multiply and divide by

9

= {7 \over 9}\left[ {{9 \over {10}} + {{99} \over {100}} + {{999} \over {1000}} + ......} \right.\,\,

$+$ up to

20

terms ]

= {7 \over 9}\left[ {\left( {1 - {1 \over {10}}} \right)} \right. + \left( {1 - {1 \over {{{10}^2}}}} \right) + \left( {1 - {1 \over {{{10}^3}}}} \right) + ......

$+$ up to

20

terms ]

= {7 \over 9}\left[ {20 - {{{1 \over {10}}\left( {1 - {{\left( {{1 \over {10}}} \right)}^{20}}} \right)} \over {1 - {1 \over {10}}}}} \right]

= {7 \over 9}\left[ {{{179} \over 9} + {1 \over 9}{{\left( {{1 \over {10}}} \right)}^{20}}} \right]

= {7 \over {81}}\left[ {179 + {{\left( {10} \right)}^{ - 20}}} \right]

Q89

If a1, a2, a3, ............... an are in A.P. and a1 + a4 + a7 + ........... + a16 = 114, then a1 + a6 + a11 + a16 is equal to :

A 38

B 98

C 76

D 64

Correct Answer

Option C

Solution

3(a1 + a16) = 114

{a_1} + {a_{16}} = 38

Now a1 + a6 + a11 + a16 = 2(a1 + a16) = 2 × 38 = 76

Q90

Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is

A 5

B 3/5

C 8/5

D 1/5

Correct Answer

Option B

Solution

Let

a=

first team of

G.P.

and

r=

common ratio of

G.P.;

Then

G.P.

is

a,

ar,

a{r^2}

Given

{S_\infty } = 20 \Rightarrow {a \over {1 - r}} = 20

\Rightarrow a = 20\left( {1 - r} \right)....\left( i \right)

Also

{a^2} + {a^2}{r^2} + {a^2}{r^4} + ...

to

\infty = 100

\Rightarrow {{{a^2}} \over {1 - {r^2}}} = 100

\Rightarrow {a^2} = 100\left( {1 - r} \right)\left( {1 + r} \right)....\left( {ii} \right)

From

(i),

{a^2} = 400{\left( {1 - r} \right)^2};

From

(ii),

we get

100\left( {1 - r} \right)\left( {1 + r} \right)

\,\,\,\,\,\,\,\,\,\, = 400{\left( {1 - r} \right)^2}

\Rightarrow 1 + r = 4 - 4r

\Rightarrow 5r = 3

\Rightarrow r = 3/5.