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

\(x=\frac{1}{5+\frac{1}{6+x}}\)

\(\Rightarrow x=\frac{6+x}{31+5 x}\)

\(\Rightarrow 5 x^{2}+31 x=x+6\)

\(\Rightarrow 5 x^{2}+30 x-6=0\)

\(\Rightarrow x=\frac{-30 \pm \sqrt{900+120}}{10}=-3 \pm \sqrt{10.2}\)

The continued fraction has to be positive.

Therefore, We reject the negative value.

Concept:

A series in which the terms are alternatively positive or negative is called an alternating series.

Liebnitz’s series:

An alternating series u₁ – u₂ + u₃ – u₄ + … converges if

(i) Each term is numerically less than its preceding term,

(ii) \(\mathop {\lim }\limits_{n \to \infty } {u_n} = 0\)

If \(\mathop {\lim }\limits_{n \to \infty } {u_n} \ne 0\), the given series is oscillatory.

Calculation:

Given series is \(\mathop \sum \limits_{n = 1}^\infty \frac{{{{\left( { - 1} \right)}^{n - 1}}n}}{{5n - 1}}\)

Now \({u_n} = \frac{n}{{5n - 1}} \)

⇒ \({u_n} - {u_{n - 1}} = \frac{n}{{5n - 1}} - \frac{{n - 1}}{{5n - 8}} = \frac{{ - 2n-1}}{{\left( {5n - 1} \right)\left( {5n - 8} \right)}}\) < 0

So each term is numerically less than its preceding term.

Now limit,

\(\mathop {\lim }\limits_{n \to \infty } {u_n} = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{5n - 1}} = \frac{1}{2}\)

⇒ \(\mathop {\lim }\limits_{n \to \infty } {u_n} \ne 0\)

∴ Series is oscillatory

Given:

\(\rm \Sigma \frac{\sqrt{5n^2-5n+1}}{7n^3-7n^2+2} \)

Concept used:

Limit Comparision test:

if an and bn are two positive series such that \(\underset{n \rightarrow \infty}{L t} \frac{a_n}{b_n} = c \) 

where c > 0 and finite then, either Both series converges or diverges together

P - Series test: 

∑ \(\frac{1}{n^p }\)is convergent for p > 1 and divergent for p ≤  1

Calculations:

n^th term of the given series = u_n = \(\rm \Sigma \frac{\sqrt{5n^2-5n+1}}{7n^3-7n^2+2} \)

Let \(\rm v_n=\frac{1}{n^2}\)

\(\rm \displaystyle Lt_{n\rightarrow\infty}\frac{u_n}{v_n}=\displaystyle Lt_{n\rightarrow\infty}\left[\frac{n\sqrt{5-\frac{5}{n}+\frac{1}{n^2}}}{n^3\left(7-\frac{7}{n}+\frac{2}{n^3}\right)}\times\frac{n^2}{1}\right] \)

\(\rm =\displaystyle Lt_{n\rightarrow\infty}\left[\frac{\sqrt{5-\frac{5}{n}+\frac{1}{n^2}}}{\left(7-\frac{7}{n}+\frac{2}{n^3}\right)}\right]=\frac{\sqrt5}{7}\ne0 \)

∴ By comparison test, Σu_n and Σv_n both converge or diverge.

But Σv_n is convergent. [p series test  - p = 2 > 1]

 ∴ Σu_n is convergent.

Given:

\(\rm \displaystyle \Sigma_{n=1}^{\infty}\left(\frac{5^n+3}{6^n+1}\right)^{1/2} \)

Concept used:

Limit Comparision test:

if an and bn are two positive series such that \(\underset{n \rightarrow \infty}{L t} \frac{a_n}{b_n} = c \)  where c > 0 and finite then, either Both series converge or diverge together

P - Series test: 

∑ \(\frac{1}{n^p }\)is convergent for p > 1 and divergent for p ≤  1

\(\rm \frac{u_n}{v_n}=\left(\frac{1+\frac{3}{5^n}}{1+\frac{1}{6^n}}\right)^{1/2} \)

\(\rm \displaystyle Lt_{n\rightarrow \infty}\frac{u_n}{v_n}=1\ne0 \) ;

∴ By comparison test, Σu_n and Σv_n behave the same way.

But Σv_n = \(\rm Σ_{n=1}^{\infty}\left(\frac{5}{6}\right)^{n/2}=\sqrt{\frac{5}{6}}+\frac{5}{6}+\left(\frac{5}{6}\right)^{3/2}+.... \) which is a geometric series with common ratio \(\rm \sqrt{\frac{5}{6}}\) which is less than 1.

∴ Σv_n is convergent.

Hence Σu_n is convergent.

Calculation

α = 1² + 4² + 8² …. 

⇒ t_n = an² + bn + c 

⇒ 1 = a + b + c 

⇒ 4 = 4a + 2b + c 

⇒ 8 = 9a + 3b + c 

On solving we get, \(\mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{3}{2}, \mathrm{c}=-1\)

⇒ \(\alpha=\sum_{n=1}^{10}\left(\frac{n^2}{2}+\frac{3 n}{2}-1\right)^2\)

⇒ \(4 \alpha=\sum_{n=1}^{10}\left(n^2+3 n-2\right)^2, \beta=\sum_{n=1}^{10} n^4 \)

⇒ \(4 \alpha-\beta=\sum_{n=1}^{10}\left(6 n^3+5 n^2-12 n+4\right)=55(353)+40\)

On comparing

⇒  k = 353

Concept:

a + ar + ar2 + ar3 +….. 

Sum of the above infinite geometric series:

\(=\frac{a}{1-r}\)

Analysis:

Given:

1 + 2(a2 + 1) + 3(a2 + 1)2 + 4(a2 + 1)3 + ......

let x = (a2 + 1)

The series now becomes

S = 1 + 2x + 3x² + 4x³ + ......  ----(1)

By multiplying x on both sides we get

xS = x + 2x² + 3x³ + 4x⁴ + ...... ----(2)

Subtracting (1) and (2), we get

S(1 - x) = 1 + x + x² + x³ + ..... ---(3)

The right hand side of (3) forms infinite geometric series with a = 1, r = x

∴ S(1 - x) = \(\frac{1}{1-x}\)

\(\Rightarrow S = \frac{1}{(1-x)^2}\)

putting the value of x, we get

\(\Rightarrow S = \frac{1}{(1- a^2 - 1)^2}\)

\(\Rightarrow S = \frac{1}{a^4}\)

The pattern followed here is –

Hence 16 will complete the series.

Concept:

The N^th term test

If \(\lim_{n\rightarrow ∞ }\left ( \sum_{n=0}^{∞ }a_{n} \right )=L\), where L is any tangible number other than zero. Then, \(\left ( \sum_{n=0}^{∞ }a_{n} \right )\) diverges.

This is also called the Divergence test.

Calculation:

We have, \(\left\)

\(\Rightarrow \sum_{n=1}^{∞ } log\left ( \frac{1}{n} \right )\)

\(\Rightarrow \lim_{n \to ∞ }\left [\sum_{n=1}^{∞ } log\left ( \frac{1}{n} \right ) \right ]\)

\(\Rightarrow \lim_{n \to ∞ }log\left ( \frac{1}{n} \right )\)

So, as n → ∞, \(\frac{1}{n}\) → 0

\(\Rightarrow \lim_{n \to ∞ }log\left ( \frac{1}{n} \right ) = -∞ \neq 0\)

Thus, our series diverges to -∞ by the n^th term test.

Hence, The sequence \(\left\) is divergent to -∞.

Let the given expansion be f(n)

f(n) = (1 + a1)(1 + a2)... (1 + an)

Also given, a1 = a2 = a3 = ... = an = 1

Consider for n = 2

f(2) = (1 + a1)(1 + a2)

f(2) = (1 + 1)(1 + 1) = 2²

Consider for n = 5

f(5) = (1 + a1)(1 + a2)(1 + a3)(1 + a4)(1 + a5)

f(5) = (1 + 1)(1 + 1)(1 + 1)(1 + 1)(1 + 1) = 25

Similary for n times, it is given as

f(n) = (1 + 1)(1 + 1) ...... (1 + 1) = 2ⁿ

(1 + a1)(1 + a2)... (1 + an) = 2n

Concept Used:-

If the summation from 0 to n such that \(\rm\displaystyle\sum_{r = 0}^{ n}\) ar xr is given, then the expanded form of it can be written as,

 ⇒ \(\rm\displaystyle\sum_{r = 0}^{ n}\) ar xr = a₀+a₁ x+a₂ x²+a₃ x³+a₄ x⁴+..........+a_nxⁿ

Explanation:-

Given,

(1+ x + x2)n = \(\rm\displaystyle\sum_{r = 0}^{2 n}\) ar xr, 

On the expanding right-hand side, we get,

\(\left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+\cdots+a_{2 n} x^{2 n}\)

Differentiate it with respect to x,

\(n\left(1+x+x^2\right)^{n-1}(1+2 x)=a_1+2 a_2 x+3 a_3 x^2+4 a_4 x^3+\cdots+2 n a_n x^{2 x-1}\)Now put x = -1 in the above equation,

\(\Rightarrow n(1-1+1)^{n-1}(1-2)=a_1-2 a_2+3 a_3-4 a_4+5a_5 \cdots-2 n a_{2n}\\ \Rightarrow n^1(1)^{n-1}(-1)=a_1-2 a_2+3 a_3-4 a_4+5a_5 \cdots-2 n a_{2n}\\ \Rightarrow -n=a_1-2 a_2+3 a_3-4 a_4+5a_5 \cdots-2 n a_{2n}\)

So, the value of a1 − 2a2 + 3a3 − …. −2n a2n is -n.

Hence, the correct option is 3.

Concept Used:

Ratio Test:

Let \( L=\lim _{n \rightarrow ∞} \frac{\left|a_{n+1}\right|}{\left|a_n\right|}\)

• If L < 1, then \(\sum a_n\) converges absolutely.
• If L > 1, or the limit goes to ∞ , then \(\sum a_n\) diverges.
• If L = 1 or if L does not exist, then the test fails, and we know nothing.

Calculation:

\( \left( {\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{{{\rm{x}}^{{\rm{n - 1}}}}}}{{{\rm{n}}\left( {{{\rm{3}}^{\rm{n}}}} \right)}}} } \right) \) = \( \left( {\sum_{n=0}^\infty a_n} \right) \)

\( a_n=\frac{x^{n-1}}{n(3^n)}\) and \( a_{n+1}=\frac{x^{n}}{(n+1)3^{n+1}}\)

⇒ \(L=\lim_{n\to \infty}\frac{\frac{x^{n}}{(n+1)3^{n+1}}}{\frac{x^{n-1}}{n(3^n)}}\)

⇒ \(L=\lim_{n\to \infty} \frac{x}{3} \cdot \frac{n}{n+1}\)

⇒ \( L=\lim_{n\to \infty} \frac{x}{3} \cdot \left(1-\frac{1}{n+1}\right) \)

⇒ \(L=\frac{x}{3}\)

For Convergent, \( \frac{x}{3} <1\)

⇒ x < 3

It is a MSQ question.

Correct answers would be option(2) and option (4)

Concept:

To check whether the given expression converges or not we will apply Ration test and P test.

Ratio Test:

Let an is the given expression.

Let \(L=\lim_{n \mapsto \infty }|\frac{a_{n+1}}{a_{n}}|\)

if L<1, then the series Converges

if L>1, then the series Diverges

if L<1, then the test is Inconclusive.

P series test:

 \(\sum_{n=1}^{\infty }\frac{1}{n^{p}}= \frac{1}{1^{p}}+\frac{1}{2^{p}}+\frac{1}{3^{p}}+..... \) where p> 0 by definition

If p > 1 , then the series converges. If 0 < p <= 1 , then the series diverges.

Calculation:

Option (1): Substitute c= 1, d= -1

Perform  test:

\(\sum_{n=1}^{\infty }\frac{n^{d}}{c^{n}}\\ =\sum_{n=1}^{\infty }\frac{n^{-1}}{1^{n}}\\ =\sum_{n=1}^{\infty }\frac{1}{n}\)   

As, p=1 , hence series will diverge.

Hence, option (1) is incorrect.

Option (2):

Substitute c=2, d= 1.

Perform Ration test

\( \lim_{n \mapsto \infty }|\frac{a_{n+1}}{a_{n}}|=\lim_{n \mapsto \infty}\frac{n+1}{2^{n+1}}\times \frac{2^{n}}{n}=\frac{1}{2}\)

​​​​As, L < 1 , the series converges.

Hence, option (2) is correct 

Option (3): 

​Substitute c= 0.5, d= -10

Perform Ratio test;

\( \sum a_{n}= \sum \frac{n^{-10}}{(0.5)^{n}}\\ \lim_{n \mapsto \infty }|\frac{a_{n+1}}{a_{n}}|=\lim_{n \mapsto \infty}\frac{(n+1)^{-10}}{(0.5)^{n+1}}\times \frac{0.5^{n}}{n^{10}}=\frac{1}{0.5}=2\)

As L>1, the series will diverge.

Hence, option (3) is incorrect.

Option (4):

​Substitute c= 1, d= -2

Perform P test:

• \(\sum a_{n}= \sum \frac{n^{-2}}{(1)^{n}}=\sum\frac{1}{n^{2}}\)
• Here, p > 1 , therefore the series converges.
• Hence, option (4) is correct.

Concept:

The comparison test in the limit form,

Consider two positive term series \(\mathop \sum \limits_{n = 1}^\infty {a_n}\) and \(\mathop \sum \limits_{n = 1}^\infty {b_n}\) such that \(\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = L\) (finite), then both the series either converge or diverge.

Calculation:

Given series is

\(\mathop \sum \limits_{n = 1}^\infty \sqrt {\frac{{{5^n} + 1}}{{{3^n} - 1}}} \)

⇒ \({a_n} = \sqrt {\frac{{{5^n} + 1}}{{{3^n} - 1}}} = {\left( {\frac{5}{2}} \right)^{\frac{n}{2}}}\sqrt {\frac{{1 + \frac{1}{{{5^n}}}}}{{1 - \frac{1}{{{3^n}}}}}} \)

Lets take \({b_n} = {\left( {\frac{5}{2}} \right)^{\frac{n}{2}}}\) 

By comparison test in limit form,

\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{a_n}}}{{{b_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \sqrt {\frac{{1 + \frac{1}{{{5^n}}}}}{{1 - \frac{1}{{{3^n}}}}}} = 1\;\left( {finite} \right)\)

Therefore, both the series either converge or diverge.

The series ∑b_n is a geometric progression with r = 2.5, since r > 1, series is divergent.

As b_n is divergent, a_n is also divergent.

Concept:

a + ar + ar2 + ar3 +….. 

Sum of the above infinite geometric series:

\(=\frac{a}{1-r}\)

Analysis:

Given:

1 + 2(a2 + 1) + 3(a2 + 1)2 + 4(a2 + 1)3 + ......

let x = (a2 + 1)

The series now becomes

S = 1 + 2x + 3x² + 4x³ + ......  ----(1)

By multiplying x on both sides we get

xS = x + 2x² + 3x³ + 4x⁴ + ...... ----(2)

Subtracting (1) and (2), we get

S(1 - x) = 1 + x + x² + x³ + ..... ---(3)

The right hand side of (3) forms infinite geometric series with a = 1, r = x

∴ S(1 - x) = \(\frac{1}{1-x}\)

\(\Rightarrow S = \frac{1}{(1-x)^2}\)

putting the value of x, we get

\(\Rightarrow S = \frac{1}{(1- a^2 - 1)^2}\)

\(\Rightarrow S = \frac{1}{a^4}\)

Concept:

A p-series is a specific type of infinite series. It's a series of the form as shown below,

\(\mathop \sum \nolimits_{n = 1}^\infty \frac{1}{{{n^p}}} = \frac{1}{{{1^p}}}+\frac{1}{{{2^p}}}+\frac{1}{{{3^p}}}+... \)

With p-series,

If p > 1, the series will converge, If 0 < p ≤ 1, the series will diverge.

The Ratio test:

Theorem: Let ∑an be a series and If \(\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{k + 1}}}}{{{a_k}}} = L\), and

if L < 1 then the series converges, but if L > 1 the series diverges.

If L = 1, then the test is inconclusive, then we have to go for another test.

Explanation:

Consider the first series:

 \(\mathop \sum \limits_{n = 1}^\infty {a_n} \Rightarrow \mathop \sum \limits_{n = 1}^\infty \frac{1}{{n\sqrt n }} \Rightarrow \mathop \sum \limits_{n = 1}^\infty \frac{1}{{{n^{\frac{3}{2}}}}}\)

This takes the form of p series \(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{{n^p}}}\)

Here p = 3/2

p > 1 ⇒ Convergent

Consider the second series \(\mathop \sum \limits_{n = 1}^\infty {b_n} \Rightarrow \mathop \sum \limits_{n = 1}^\infty \frac{1}{{n!}}\)

By ratio test, \( \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{{b_{n + 1}}}}{{{b_n}}} = {\rm{lim}}\frac{1}{{n + 1}} = 0\)

Here L = 0 (< 1) ⇒ Convergent

Therefore, both the series are convergent.