Solution & Explanation

### Related Formula Partial fraction decomposition for telescopic summing:

frac2n^2 - (n+1)^2 + 1 dotsdots text structures directly cancel in series expansions.

### Core Logic Given recurrence:

a_n+1 - frac12a_n = fracn^2 - 2n - 1n^2(n+1)^2

Rewrite the numerator to split the fraction:

n^2 - 2n - 1 = 2n^2 - (n^2 + 2n + 1) = 2n^2 - (n+1)^2

a_n+1 - frac12a_n = frac2n^2 - (n+1)^2n^2(n+1)^2 = frac2(n+1)^2 - frac1n^2

### Step 1: Telescope generation Multiply both sides by appropriate powers of 2 to create a cancelling chain: For

n=1

a_2 - frac12a_1 = frac22^2 - frac11^2

For

n=2

: multiply by 2

Rightarrow 2left[a_3 - frac12a_2 = frac23^2 - frac12^2right] Rightarrow 2a_3 - a_2 = frac2 times 23^2 - frac22^2

Wait, let's look at a cleaner telescopic scaling:

a_n+1 - frac2(n+1)^2 = frac12 left(a_n - frac2n^2right)

. ### Step 2: Identify Geometric Progression Let

V_n = a_n - frac2n^2

. The recurrence gives

V_n+1 = frac12 V_n

. This proves

V_n

is a geometric progression with common ratio

r = 1/2

. First term

V_1 = a_1 - frac21^2 = 1 - 2 = -1

. ### Step 3: Infinite Summation We need

left| sum_n=1^infty left( a_n - frac2n^2 right) right| = left| sum_n=1^infty V_n right|

. Since

V_n

is an infinite GP:

S_infty = fracV_11 - r = frac-11 - 1/2 = frac-11/2 = -2

Taking absolute value:

|-2| = 2

### Pattern Recognition When dealing with rational fraction recurrences

A_n+1 - k A_n = f(n) - k f(n-1)

, immediately substitute

V_n = A_n - f(n)

. This substitution instantly isolates a classical Geometric Progression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

More Sequences and Series Previous-Year Questions — Page 2

Q70 jee_main_2025_03_april_evening Special Series

The sum

1 + frac1+32! + frac1+3+53! + frac1+3+5+74! + dots

is equal to

A. $6e$ $6e$
B. $4e$ $4e$
C. $3e$ $3e$
D. $2e$ $2e$

Solution

### Related Formula Sum of first

r

odd natural numbers:

sum_k=1^r (2k-1) = r^2

Exponential series expansion:

sum_r=0^infty frac1r! = e

### Core Logic Let's find the general

r

-th term of the series:

T_r = frac1 + 3 + 5 + dots + (2r-1)r! = fracr^2r! = fracr(r-1)!

### Step 1: Expressing term in terms of sum limits Let's write

r = (r-1) + 1

T_r = fracr-1+1(r-1)! = frac1(r-2)! + frac1(r-1)!

Our infinite sum is:

S = sum_r=1^infty T_r = sum_r=2^infty frac1(r-2)! + sum_r=1^infty frac1(r-1)!

Both sums are standard representations of the exponential expansion. ### Step 2: Summing the parts - First part:

sum_r=2^infty frac1(r-2)! = 1 + frac11! + frac12! + dots = e

- Second part:

sum_r=1^infty frac1(r-1)! = 1 + frac11! + frac12! + dots = e

textTotal Sum S = e + e = 2e

### Pattern Recognition The general term containing

r^2

in summation with factorials converges to

2e

. Remember the shortcut:

sum fracr^2r! = 2e

sum fracr^3r! = 5e

. It is extremely useful to memorize these common limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series Class 12 Mathematics: Limits, Continuity and Differentiability

Q63 jee_main_2025_07_april_morning Arithmetico-Geometric Progression

Let

mathbfx_1, mathbfx_2, mathbfx_3, mathbfx_4

be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from

mathbfx_1, mathbfx_2, mathbfx_3, mathbfx_4

then the resulting numbers are in an arithmetic progression. Then the value of

frac124 (mathbfx_1 mathbfx_2 mathbfx_3 mathbfx_4)

is:

A. $72$ $72$
B. $18$ $18$
C. $36$ $36$
D. $216$ $216$

Solution

### Related Formula For a geometric progression, the terms can be set as

a, ar, ar^2, ar^3

. For three terms

A, B, C

to be in arithmetic progression, they must satisfy:

2B = A + C

### Core Logic Let the elements be

x_1 = a, x_2 = ar, x_3 = ar^2, x_4 = ar^3

. After the specified subtractions, the sequence becomes:

a - 2, quad ar - 7, quad ar^2 - 9, quad ar^3 - 5

Since this sequence is in AP, we form two separate common difference linear linkages:

2(ar - 7) = (a - 2) + (ar^2 - 9) implies 2ar - 14 = ar^2 + a - 11 implies ar^2 - 2ar + a + 3 = 0 quad dots (1)

2(ar^2 - 9) = (ar - 7) + (ar^3 - 5) implies 2ar^2 - 18 = ar^3 + ar - 12 implies ar^3 - 2ar^2 + ar + 6 = 0 quad dots (2)

### Step 1: Solve the Simultaneous Polynomials Multiply equation (1) by

r

ar^3 - 2ar^2 + ar + 3r = 0 quad dots (3)

Subtract equation (3) from equation (2):

(ar^3 - 2ar^2 + ar + 6) - (ar^3 - 2ar^2 + ar + 3r) = 0

6 - 3r = 0 implies 3r = 6 implies r = 2

Substitute

r = 2

back into equation (1):

a(2)^2 - 2a(2) + a + 3 = 0

4a - 4a + a + 3 = 0 implies a = -3

### Step 2: Find the Continuous Product Value The continuous product term is:

mathbfx_1mathbfx_2mathbfx_3mathbfx_4 = a cdot ar cdot ar^2 cdot ar^3 = a^4 r^6

mathbfx_1mathbfx_2mathbfx_3mathbfx_4 = (-3)^4 cdot (2)^6 = 81 times 64 = 5184

Now divide by 24 as required:

frac124(5184) = 216

### Pattern Recognition Notice that multiplying the first AP condition equation by

r

perfectly mimics the structure of the second condition equation except for the absolute scalar value, allowing direct elimination of all polynomial variable indices simultaneously. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

Q55 jee_main_2025_08_april_evening Infinite Series

frac11^4 + frac12^4 + frac13^4 + dots infty = fracpi^490

, and

frac11^4 + frac13^4 + frac15^4 + dots infty = alpha

frac12^4 + frac14^4 + frac16^4 + dots infty = beta

then

fracalphabeta

is equal to

A. $23$ $23$
B. $18$ $18$
C. $15$ $15$
D. $14$ $14$

Solution

### Related Formula

textTotal Sum = alpha + beta

### Core Logic Factor out common fractions from the even terms component (

beta

) to represent it as a scalar multiple of the universal sum sequence. ### Step 1: Simplify the Even Terms Series

beta = frac12^4 + frac14^4 + frac16^4 + dots = frac12^4 left( frac11^4 + frac12^4 + frac13^4 + dots right)

beta = frac116 left( fracpi^490 right)

### Step 2: Express Alpha by Remainder Deduction Since total sum equals

alpha + beta

alpha = textTotal Sum - beta = fracpi^490 - frac116 left( fracpi^490 right) = frac1516 left( fracpi^490 right)

### Step 3: Compute the Relative Ratio

決fracalphabeta = fracfrac1516 left( fracpi^490 right)frac116 left( fracpi^490 right) = 15

### Pattern Recognition For alternating p-series powers like

\sum n^{-p}

, the even component fractions always condense via factor steps to

2^{-p} \cdot S_{\text{total}}$, decoupling power values cleanly from final simple scalar quotients. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

Q57 jee_main_2025_29_jan_evening Infinite Geometric Progression

Let

mathbfS = mathbfN cup \0\

. Define a relation \mathbf{R} from

mathbfS

mathbfR

by:

mathbfR = left\left(x, yright): log_e y = x log_e left(frac25right), x in S, y in R right\.

Then, the sum of all the elements in the range of

mathbfR

is equal to

A. $frac32$ $\frac{3}{2}$
B. $frac53$ $\frac{5}{3}$
C. $frac109$ $\frac{10}{9}$
D. $frac52$ $\frac{5}{2}$

Solution

### Related Formula Sum of an infinite geometric progression with

|r| < 1

S_infty = fraca1 - r

### Core Logic From the definition of the relation:

log_e y = x log_eleft(frac25 ight) implies log_e y = log_eleft(frac25right)^x implies y = left(frac25 ight)^x

Infinite Geometric Progression diagram for Q57 - JEE Main 2025 Evening

Since

x in S = \0, 1, 2, 3, dots\

, the output values of

y

represent elements of the range. ### Step 1: Compute Infinite Sum Generating elements by plugging in values of

x

: For

x = 0 implies y = 1

For

x = 1 implies y = frac25

For

x = 2 implies y = left(frac25 ight)^2

Sum of elements in the range:

textSum = 1 + left(frac25 ight)^1 + left(frac25 ight)^2 + dots = frac11 - frac25 = frac53

### Pattern Recognition Convert log equations into standard exponential equations right away. A variable index belonging to whole numbers indicates an infinite GP summation scenario. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series Class 11 Mathematics: Relations and Functions

Q73 jee_main_2025_29_jan_evening Arithmetic Progression Properties

Let

a_1, a_2, ldots, a_2024

be an Arithmetic Progression such that

a_1 + (a_5 + a_10 + a_15 + ldots + a_2020) + a_2024 = 2233

. Then

a_1 + a_2 + a_3 + ldots + a_2024

is equal to

Numerical Answer. Answer: 11132 to 11132

Solution

### Related Formula Symmetry identity rule inside Arithmetic Progressions:

a_k + a_n-k+1 = a_1 + a_n

### Core Logic Group matching paired steps equidistant from sequence boundary ends:

a_1 + a_2024 = a_5 + a_2020 = a_10 + a_2015 = dots

The sequence of inner indices follows an AP tracking loop:

5, 10, 15, dots, 2020

Calculate internal block element count

N

2020 = 5 + (N-1)5 implies 2015 = 5(N-1) implies N - 1 = 403 implies N = 404 text terms

### Step 1: Simplify Expression Equations Since the inner sequence contains 404 terms, they form exactly 202 symmetrical pairs. Adding

a_1

and

a_2024

introduces one more pair, resulting in 203 identical sum blocks:

203(a_1 + a_2024) = 2233 implies a_1 + a_2024 = frac2233203 = 11

### Step 2: Evaluate the Total Sum Using the standard AP sum formula:

S_2024 = frac20242(a_1 + a_2024) = 1012 times 11 = 11132

### Pattern Recognition Progressions possess natural positional balance. Grouping symmetrical index pairs (

a_k + a_n-k+1

) allows factoring out variable steps right away. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

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Let $a_1 = 1$ $a_1 = 1$ and for $n geq 1$ $n \geq 1$ , $a_n+1 = frac12 a_n + fracn^2 - 2n - 1n^2 (n + 1)^2$ . Then $left|sum_n=1^inftyleft(a_n - frac2n^2right)right|$ is equal

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Let a_1 = 1$a_1 = 1$ and for n geq 1$n \geq 1$ , a_n+1 = frac12 a_n + fracn^2 - 2n - 1n^2 (n + 1)^2$a_{n+1} = \frac{1}{2} a_n + \frac{n^2 - 2n - 1}{n^2 (n + 1)^2}$ . Then left|sum_n=1^inftyleft(a_n - frac2n^2right)right|$\left|\sum_{n=1}^{\infty}\left(a_n - \frac{2}{n^2}\right)\right|$ is equal

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Let $a_1 = 1$ $a_1 = 1$ and for $n geq 1$ $n \geq 1$ , $a_n+1 = frac12 a_n + fracn^2 - 2n - 1n^2 (n + 1)^2$ . Then $left|sum_n=1^inftyleft(a_n - frac2n^2right)right|$ is equal