Summation MCQ Quiz - Objective Question with Answer for Summation - 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

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 \(\lim_{n\rightarrow ∞ }\left ( \sum_{n=0}^{∞ }a_{n} \right )=L\), 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 -∞.

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

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