Special Series MCQ Quiz - Objective Question with Answer for Special Series - Download Free PDF

Given:

The sequence P1, P2, P3, ..... is defined by P1 = 211, P2 = 375, P3 = 420, P4 = 523, and Pn = Pn-1 - Pn-2 + Pn-3 - Pn-4 for all n ≥ 5.

Calculation:

We need to find the value of P531 + P753 + P975.

Given the recurrence relation Pn = Pn-1 - Pn-2 + Pn-3 - Pn-4, we notice that this sequence is periodic with a period of 4. This means that every 4 terms, the sequence repeats itself.

Let's calculate the first few terms to identify the pattern:

P1 = 211

P2 = 375

P3 = 420

P4 = 523

P5 = P4 - P3 + P2 - P1 = 523 - 420 + 375 - 211 = 267

P6 = P5 - P4 + P3 - P2 = 267 - 523 + 420 - 375 = -211

P7 = P6 - P5 + P4 - P3 = -211 - 267 + 523 - 420 = -375

P8 = P7 - P6 + P5 - P4 = -375 - (-211) + 267 - 523 = -420

P9 = P8 - P7 + P6 - P5 = -420 - (-375) + (-211) - 267 = -523

We can observe the periodicity:

P1 = 211

P2 = 375

P3 = 420

P4 = 523

P5 = 267

P6 = -211

P7 = -375

P8 = -420

P9 = -523

P10 = 267

The sequence repeats every 4 terms: P5 = 267, P6 = -211, P7 = -375, P8 = -420, P9 = -523, P10 = 267, and so on.

To find P531, P753, and P975, we need to determine their positions in the cycle:

P531 = P(531 mod 4) = P3 = 420

P753 = P(753 mod 4) = P1 = 211

P975 = P(975 mod 4) = P3 = 420

Therefore, P531 + P753 + P975 = 420 + 211 + 420 = 1051

The correct answer is option 1) 898.

Concept:

If a1, a2, a3,...are in GP with common ratio r, 

 \(​​​​r=\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\frac{a_{4}}{a_{3}}\)

If a be the first term, r be the common ratio of a GP then,

 \(\rm S_n = \frac{a(1-r^n)}{1-r}\)

Calculation:

Calculation:

Let required sum is S.

⇒ S = \(\frac{1}{2} + \frac{3}{4}+\frac{7}{8}+\frac{15}{16}+...\)

⇒ S =\(\frac{2-1}{2} + \frac{4-1}{4}+\frac{8-1}{8}+\frac{16-1}{16}+...\)

⇒ S = \(1-\frac{1}{2} +1- \frac{1}{4}+1-\frac{1}{8}+1-\frac{1}{16}+...n\ term\)

⇒ S = \( n-[\frac{1}{2} + \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...n\ term]\)

Threfore, a = 1/2 and r = 1/2

We know that the sum of the n^th term of GP (r < 1) is 

\(\rm S_n = \frac{a(1-r^n)}{1-r}\)

⇒ S = \(n- \frac{(1/2)[1-(1/2)^n]}{1-1/2}\)

⇒ S = \(n- \frac{(1/2)[1-(2)^{-n}]}{1/2}\)          (∵ 1/aⁿ = a^-n)

⇒ S = n - 1 + 2-n

∴ S = n + 2-n -1

Concept:

1. The nth term of the A.P. is given by:

an = a + (n – 1) d

2. The sum of n terms of an AP with first term a and common difference d is given by: 

\(\rm {S_n} = \frac{n}{2} × \left[ {2a + \left( {n - 1} \right)d} \right]\)      ----(1)

Or

\({S_n} = \frac{n}{2} × \left[ {a + l} \right]\)

Where,

an = nth term and l = Last term

Calculation:

Given: 

\(3 + \dfrac{9}{2} + 6 + \dfrac{15}{2} + .....\)

d = 3/2

n = 25

a = 3

From equation (1);

\(S_{25} = \dfrac{25}{2} [2a + (25-1) \times d] \)

\(= \dfrac{25}{2} \left[ 2 \times 3 + (24) \times \dfrac{3}{2} \right]\)

\(= \dfrac{25}{2} \times 42 = 25 \times 21 = 525\)

Concept: 

Sum of consecutive numbers from 1 to n: \( \rm 1 + 2 + 3 +\ ...\ + n = \dfrac{n(n+1)}{2}\).

Calculation:

The sum of the first 24 terms of the given series can be written as:

\(\rm \sqrt{2}+ \sqrt{8}+\sqrt{18}+\sqrt{32}+\ ...\)

\(\rm =\sqrt{2}+ 2\sqrt{2}+3\sqrt{2}+4\sqrt{2}+\ ...\ +24\sqrt2\)

\(\rm =\sqrt2(1+2+\ ...\ +24)\)

\(\rm =\sqrt2\times \dfrac{24\times25}{2}=300\sqrt2\).

Calculation:

Given:

Series: \(\frac{1}{2 \times 5} + \frac{1}{5 \times 8} + \frac{1}{8 \times 11} + \dots + \frac{1}{(3n-1)(3n+2)}\)

\(\frac{1}{(3n-1)(3n+2)} = \frac{A}{3n-1} + \frac{B}{3n+2}\)

⇒ \(1 = A(3n+2) + B(3n-1)\)

Let n = 1/3: 1 = A(3) => A = 1/3

Let n = -2/3: 1 = B(-3) => B = -1/3

⇒ \(\frac{1}{(3n-1)(3n+2)} = \frac{1}{3} \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right)\)

series: \(\frac{1}{3} \left( \frac{1}{2} - \frac{1}{5} \right) + \frac{1}{3} \left( \frac{1}{5} - \frac{1}{8} \right) + \frac{1}{3} \left( \frac{1}{8} - \frac{1}{11} \right) + \dots + \frac{1}{3} \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right)\)

⇒ \(\frac{1}{3} \left( \frac{1}{2} - \frac{1}{5} + \frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{11} + \dots + \frac{1}{3n-1} - \frac{1}{3n+2} \right)\)

⇒ \(\frac{1}{3} \left( \frac{1}{2} - \frac{1}{3n+2} \right)\)

⇒ \(\frac{1}{3} \left( \frac{3n+2 - 2}{2(3n+2)} \right)\)

⇒ \(\frac{1}{3} \left( \frac{3n}{2(3n+2)} \right)\)

⇒ \(\frac{n}{2(3n+2)}\)

⇒ \(\frac{n}{6n+4}\)

∴ The sum of the series is \(\frac{n}{6n+4}\)

Hence option 4 is correct.

Concept:

Expansion of ex:

\(\rm e^{x} = 1+ \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!}+ \dfrac{x^4}{4!}+ .....\)

Calculation:

\(\rm e^{x} = 1+ \dfrac{x}{1!}++ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!}+ \dfrac{x^4}{4!}+ .....\)

Put x = -1,

\(\rm e^{(-1)} = 1+ \dfrac{(-1)}{1!}++ \dfrac{(-1)^2}{2!}+ \dfrac{(-1)^3}{3!}+ \dfrac{(-1)^4}{4!}+ .....\)

\( \rm e^{-1} = 1- \dfrac{1}{1!}+ \dfrac{1}{2!}- \dfrac{1}{3!}+ \dfrac{1}{4!}- .....\)

\( \therefore \rm 1- \dfrac{1}{1!}+ \dfrac{1}{2!}- \dfrac{1}{3!}+ \dfrac{1}{4!}- ..... = \frac 1 e\)

Calculation:

\(\rm \frac{1}{{1 \times 2}} + \frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} +...+\frac{1}{{n \times (n+1)}} \)

\(\rm = \frac{{2\; - \;1}}{{1 \times 2}} + \frac{{3\; - \;2}}{{2 \times 3}} + \frac{{4\; - \;3}}{{3 \times 4}} +... + \frac{{(n+1)\; - \;n}}{{n \times (n+1)}}\)

\(\rm = \frac{1}{1} - \frac{1}{2} + \;\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} +... + \frac{1}{n} - \frac{1}{n+1}\)

\(\rm = 1 - \frac {1}{n+1}\)

\(\rm = \frac {n+1 -1}{n+1}\)

\(\rm = \frac {n}{n+1}\)

Calculation:

Given: Let S =  \(\frac{1}{{4 ⋅ 10}} + \frac{1}{{10 ⋅ 16}} + \frac{1}{{16 ⋅ 22}} + \frac{1}{{22 ⋅ 28}} + \frac{1}{{28 ⋅ 34}}\)

Now we can re-write S as shown below:

\(S = \frac{1}{6} × \left[ {\frac{{10 - 4}}{{4 ⋅ 10}} + \frac{{16 - 10}}{{10 ⋅ 16}} + \frac{{22 - 16}}{{16 ⋅ 22}} + \frac{{28 - 22}}{{22 ⋅ 28}} + \frac{{34 - 28}}{{28 ⋅ 34}}} \right]\)

\( ⇒ S = \frac{1}{6} × \left[ {\frac{1}{4} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{16}} + \ldots + \frac{1}{{28}} - \frac{1}{{34}}} \right]\)

On further simplifying the above equation we get,

⇒ S = 1/6 × [1/4 - 1/34] = 5/136

Concept:

Expansion of ex:

\(\rm e^{x} = 1+ \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!}+ \dfrac{x^4}{4!}+ .....\)

Calculation:

\(\rm e^{x} = 1+ \dfrac{x}{1!}++ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!}+ \dfrac{x^4}{4!}+ .....\)

Put x = 1,

\( \rm 1+ \dfrac{1}{1!}+ \dfrac{1}{2!}+ \dfrac{1}{3!}+ \dfrac{1}{4!}+ .....=e\)

Concept:

Arithmetic-Geometric Progression (AGP):

• The first few terms of an Arithmetic-Geometric progression composed of an arithmetic progression with first term a and common difference d, and a geometric progression with first term b and common ratio r are given by:

  ab, (a + d)br, (a + 2d)br², ... [a + (n - 1)d]br^{n - 1}, ...

• The sum to infinity of an AGP, whose |r| ≤ 1, is given by:

  S_∞ = \(\rm{ab\over1\ -\ r}\ +\ {dbr\over(1\ -\ r)^2}\)

Calculation:

The given series \({1\over2}+{3\over4}+{5\over8}+\ ...\infty\) is an AGP with

a = 1, d = 2 and b = \(1\over2\), r = \(1\over2\).

⇒ S∞ = \(\rm{1\left(1\over2\right)\over1-{1\over2}}+{(2)\left(1\over2\right)\left(1\over2\right)\over\left(1-{1\over2}\right)^2}\)

⇒ S∞  = 1 + \({1\over2}\over{1\over4}\)

⇒ S∞  = 1 + 2

⇒ S∞  = 3.

Concept:

If a₁, a₂, …., a_n be an AP then \({S_n} = {a_1} + {a_2} + \ldots + {a_n} = \frac{n}{2} \times \left( {2a + \left( {n - 1} \right)d} \right)\) where a is the 1^st term and d is the common difference.

If a₁, a₂, …, a_n be an AP then the general term is given by: a_n = a + (n - 1) × d where a is the 1^st term and d is the common difference.

Calculation:

Here, we have to find the value of 1 - 2 + 3 - 4 + 5 - ______ + 101.

⇒ 1 - 2 + 3 - 4 + 5 - ______ + 101 = (1 + 3 + …… + 101) - (2 + 4 + ….. + 100)

As, we can see that (1, 3, ……, 101) is an AP with a = 1 and d = 2.

⇒ a_n = 101 = 1 + (n - 1) × 2

⇒ n = 51

\(\Rightarrow {S_{51}} = 1 + 3 + \ldots + 101 = \frac{{51}}{2} \times \left( {2 + 50 \times 2} \right) = 2601\)

Similarly, (2, 4, …, 100) is an AP with a = 2 and d = 2

⇒ a_n = 100 = 2 + (n - 1) × 2

⇒ n = 50

\(\Rightarrow {S_{50}} = 2 + 4 + \ldots + 100 = \frac{{50}}{2} \times \left( {4 + 49 \times 2} \right) = 2550\)

⇒ 1 - 2 + 3 - 4 + 5 - ______ + 101 = (1 + 3 + …… + 101) - (2 + 4 + ….. + 100) = 2601 - 2550 = 51.

Concept: 

Sum of consecutive numbers from 1 to n: \( \rm 1 + 2 + 3 +\ ...\ + n = \dfrac{n(n+1)}{2}\).

Calculation:

The sum of the first 20 terms of the given series can be written as:

\(\rm \sqrt{5}+ \sqrt{20}+\sqrt{45}+\sqrt{80}+\ ...\)

\(\rm =\sqrt{5}+ 2\sqrt{5}+3\sqrt{5}+4\sqrt{5}+\ ...\ +20\sqrt5\)

\(\rm =\sqrt5(1+2+\ ...\ +20)\)

\(\rm =\sqrt5\times \dfrac{20\times21}{2}=210\sqrt5\).

Formula used:

n! = n × (n - 1) × (n - 2) × ....... × 3 × 2 × 1

Calculation:

a_n = n (n)!

a_n = (n + 1 - n)n!

a_n = (n + 1)n! - n!

a_n = (n + 1)! - n!

Hence, 

a₁ = 2! - 1! 

a₂ = 3! - 2! 

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

a₁₀ = 11! - 10!

Now,

 a1 + a2 + a3 +...+ a10 

2! - 1! + 3! - 2! + ......11! - 10! 

= 11! - 1

∴ The value of a1 + a2 + a3 +...+ a10 is 11! - 1.

Concept:​

• 12 + 22 + 32 + ..... + n2 = \(\rm \frac{n(n+1)(2n+1)}{6}\)

Calculation:

Here we have to find the sum of the series 2 + 8 + 18 + 32 + 50 +......+ 200 

The given series can be re-written as: 2(12 + 22 + 32 + ..... + 102)

As we know that, 12 + 22 + 32 + ..... + n2 = \(\rm \frac{n(n+1)(2n+1)}{6}\)

Hence, option 1 is the correct answer.

Concept: 

Sum of consecutive numbers from 1 to n: \(\rm1 + 2 + 3 +\ ...\ + n = \dfrac{n(n+1)}{2}\).

Calculation:

The sum of the first 18 terms of the given series can be written as:

3 + 6 + 9 + 12 + 15, ....... 

⇒ 3(1 + 2 + 3 + 4 + 5 + ....)

⇒ \(\frac{3 \times 18 \times 19}{2} = 513\)