Summation MCQ Quiz - Objective Question with Answer for Summation - Download Free PDF

Calculation:

We have,

1/1⁴ + 1/2⁴ + 1/3⁴ + 1/4⁴ + ⋯ = π⁴ / 90         (1)

This can be written as:

[1/1⁴ + 1/3⁴ + 1/5⁴ + ⋯ ] + 1/2⁴ + 1/4⁴ + ⋯ = π⁴ / 90

[1/1⁴ + 1/3⁴ + 1/5⁴ + ⋯ ] + (1/2⁴) × [1/1⁴ + 1/2⁴ + 1/3⁴ + ⋯ ] = π⁴ / 90

Using (1):

[1/1⁴ + 1/3⁴ + 1/5⁴ + ⋯ ] + (1/2⁴) × (π⁴ / 90) = π⁴ / 90

⇒ 1/1⁴ + 1/3⁴ + 1/5⁴ + ⋯ = (π⁴ / 90) × [1 − 1/2⁴] = (π⁴ / 90) × (15/16) = π⁴ / 96

Given:

First term (a) = 27

Common ratio (r) = 2/3

Find the 4th term of the GP.

Formula used:

n-th term of GP = a × r^(n-1)

Calculation:

4th term = 27 × (2/3)^(4-1)

⇒ 4th term = 27 × (2/3)³

⇒ 4th term = 27 × (8/27)

⇒ 4th term = 8

∴ The correct answer is option (1).

$x=\frac{1}{5+\frac{1}{6+x}}$

$\Rightarrow x=\frac{6+x}{31+5x}$

$\Rightarrow 5x^2+31x=x+6$

$\Rightarrow 5x^2+30x-6=0$

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

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) $\underset{n\to \mathrm{\infty }}{lim}{u}_n=0$

If $\underset{n\to \mathrm{\infty }}{lim}{u}_n\ne 0$, the given series is oscillatory.

Calculation:

Given series is $\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n-1}n}{5n-1}$

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,

$\underset{n\to \mathrm{\infty }}{lim}{u}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{n}{5n-1}=\frac{1}{2}$

⇒ $\underset{n\to \mathrm{\infty }}{lim}{u}_n\ne 0$

∴ Series is oscillatory

Given:

$\mathrm{\Sigma }\frac{\sqrt{5{\mathrm{n}}^2-5\mathrm{n}+1}}{7{\mathrm{n}}^3-7{\mathrm{n}}^2+2}$

Concept used:

Limit Comparision test:

if an and bn are two positive series such that $\underset{n\to \mathrm{\infty }}{Lt}\frac{a_n}{b_n}=c$ 

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 = $\mathrm{\Sigma }\frac{\sqrt{5{\mathrm{n}}^2-5\mathrm{n}+1}}{7{\mathrm{n}}^3-7{\mathrm{n}}^2+2}$

Let ${\mathrm{v}}_{\mathrm{n}}=\frac{1}{{\mathrm{n}}^2}$

${\displaystyle \mathrm{L}{\mathrm{t}}_{\mathrm{n}\to \mathrm{\infty }}\frac{{\mathrm{u}}_{\mathrm{n}}}{{\mathrm{v}}_{\mathrm{n}}}={\displaystyle \mathrm{L}{\mathrm{t}}_{\mathrm{n}\to \mathrm{\infty }}[\frac{\mathrm{n}\sqrt{5-\frac{5}{\mathrm{n}}+\frac{1}{{\mathrm{n}}^2}}}{{\mathrm{n}}^3(7-\frac{7}{\mathrm{n}}+\frac{2}{{\mathrm{n}}^3})}\times \frac{{\mathrm{n}}^2}{1}]}}$

$={\displaystyle \mathrm{L}{\mathrm{t}}_{\mathrm{n}\to \mathrm{\infty }}\left[\frac{\sqrt{5-\frac{5}{\mathrm{n}}+\frac{1}{{\mathrm{n}}^2}}}{(7-\frac{7}{\mathrm{n}}+\frac{2}{{\mathrm{n}}^3})}\right]=\frac{\sqrt{5}}{7}\ne 0}$

∴ 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,

The given AP is 13, 11, 9……

Formula:

T_nt = a + (n – 1)d

a = first term

d = common term

Calculation:

a = 13

d = 11 – 13

d = (-2)

T₅₀ = 13 + (50 – 1) × (-2)

⇒ T₅₀ = 13 + 49 × (-2)

⇒ T₅₀ = 13 – 98

∴ T₅₀ = -85

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

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

$\Rightarrow S=\frac{1}{a^4}$

Formula used:

Sum of A.P. = n/2{first term + last term}

Calculation:

Number of terms = n = 12

⇒ S_n = 12/2{5 + 38}

⇒ S_n = 6{43}

Given:

The first five multiples of 3

Concept:

Multiples = A multiple is a number that can be divided by another number a certain number of time without a remainder

Calculation:

⇒ The first five multiple of 3 = (3 × 1), (3 × 2), (3 × 3), (3 × 4), (3 × 5) = 3, 6, 9, 12, and 15

⇒ The sum of the multiple = 3 + 6 + 9 + 12 + 15 = 45

∴ The required result will be 45.

The pattern followed here is –

Hence 16 will complete the series.

Shortcut Trick

Formula Used:

Average = (Sum of observations)/(Number of observations)

Last term = a + (n - 1)d

Calculation: 

The above series is in arithmetic progression so the middlemost term 8th term will be the average

⇒ 8^th term = 8 + (8 - 1) × 3 = 29

⇒ Sum of the series = 29 × 15 = 435

∴ The sum of the above series is 435  

Additional Information

We can avoid this above (29 × 15) multiplication by digit sum Method and option

The digit sum of 29 is  (2 + 9) ⇒ (11) ⇒ (2) and 15 is (1 + 5) = 6 

⇒ 2 × 6 = 12 ⇒ (1 + 2) ⇒ 3 

Now check the options whose digit sum will be 3 there is only option 2 whose Digit sum is 3 

∴ 435 is the right answer

Traditional Method: 

Given:

Arithmetic progression 8 + 11 + 14 + 17 upto 15 terms

Formula Used: 

Sum of arithmetic progression = n[2a + (n - 1)d]/2

Calculation:

Sum of 1st 15 terms = 15[2 × 8 + (15-1)3]/2

⇒ (15 × 58)/2

⇒ 435

∴ 435 is the right answer

Concept:

The N^th term test

If $\underset{n\to \infty }{lim}\left(\sum _{n=0}^{\infty }{a}_n\right)=L$, where L is any tangible number other than zero. Then, $\left(\sum _{n=0}^{\infty }{a}_n\right)$ diverges.

This is also called the Divergence test.

Calculation:

We have, \(\left\)

$\Rightarrow \sum _{n=1}^{\infty }log\left(\frac{1}{n}\right)$

$\Rightarrow \underset{n\to \infty }{lim}\left[\sum _{n=1}^{\infty }log\left(\frac{1}{n}\right)\right]$

$\Rightarrow \underset{n\to \infty }{lim}log\left(\frac{1}{n}\right)$

So, as n → ∞, $\frac{1}{n}$ → 0

$\Rightarrow \underset{n\to \infty }{lim}log\left(\frac{1}{n}\right)=-\infty \ne 0$

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

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

Given:
Series 1 = $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$

Series 2 = $2+2\times \frac{1}{3}+2\times \frac{1}{3^2}+2\times \frac{1}{3^3}+...$

...

Series n = $n+n\times \frac{1}{n+1}+n\times \frac{1}{(n+1{)}^2}+n\times \frac{1}{(n+1{)}^3}+...$

Concept:
Sum of an infinite G.P. series, S_∞ = $\frac{a}{1-r}$ , when |r| < 1

Sum of an A.P. series, S_AP = $\frac{n(a+l)}{2}$ , where, $l$ = last term of series

Calculation:

For Series 1, a = 1, and r = $\frac{1}{2}$
∴ S₁ = $\frac{a}{1-r}$ = $\frac{1}{1-\frac{1}{2}}$
⇒ S₁ = 2

Similarly, for Series 2, a = 2 and r = $\frac{1}{3}$
∴ S₁ = $\frac{2}{1-\frac{1}{3}}$
⇒ S₂ = 3

Similarly,

S3 = 4

S₄ = 5, ...

S_n = n + 1

So, S₁, S₂, S₃, ... , Sn is an Arithmetic Progression (A.P.),

For A.P.,

a = S1 = 2

n = n
$l$ = Sn = n + 1,

Therefore,
(S1 + S2 + S3 + ... + S_n) = $\frac{n(a+l)}{2}=\frac{n(2+n+1)}{2}$

∴ ( S1 + S2 + S3 + ... + S_n ) = $\frac{n(n+3)}{2}$

GIVEN:

First term a = – 9

Last term l = 51

Number of term n = 16

FORMULAE USED:

Sum = n/2 × (a + l)

CALCULATION:

⇒ 16/2 × (- 9 + 51) = sum

Concept Used:-

If the summation from 0 to n such that ${\displaystyle \sum _{\mathrm{r}=0}^{\mathrm{n}}}$ ar xr is given, then the expanded form of it can be written as,

 ⇒ ${\displaystyle \sum _{\mathrm{r}=0}^{\mathrm{n}}}$ ar xr = a₀+a₁ x+a₂ x²+a₃ x³+a₄ x⁴+..........+a_nxⁿ

Explanation:-

Given,

(1+ x + x2)n = ${\displaystyle \sum _{\mathrm{r}=0}^{2\mathrm{n}}}$ ar xr, 

On the expanding right-hand side, we get,

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

Differentiate it with respect to x,

$n{(1+x+x^2)}^{n-1}(1+2x)=a_1+2a_{2}x+3a_3{x}^2+4a_4{x}^3+\cdots +2n{a}_n{x}^{2x-1}$Now put x = -1 in the above equation,

$$\begin{gathered} \Rightarrow n(1-1+1{)}^{n-1}(1-2)=a_1-2a_2+3a_3-4a_4+5a_5\cdots -2n{a}_{2n} \\ \Rightarrow n^1(1{)}^{n-1}(-1)=a_1-2a_2+3a_3-4a_4+5a_5\cdots -2n{a}_{2n} \\ \Rightarrow -n=a_1-2a_2+3a_3-4a_4+5a_5\cdots -2n{a}_{2n} \end{gathered}$$

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

Hence, the correct option is 3.