Summation MCQ Quiz in मल्याळम - Objective Question with Answer for Summation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Apr 19, 2025

നേടുക Summation ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Summation MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Summation MCQ Objective Questions

Top Summation MCQ Objective Questions

Summation Question 1:

The series un=x2n(n+1)n converges for only 

  1. x < 1 and x > 1
  2. x ≤ 1 and x ≥ 1
  3. -1 < x < 1
  4. -1 ≤ x ≤ 1

Answer (Detailed Solution Below)

Option 4 : -1 ≤ x ≤ 1

Summation Question 1 Detailed Solution

Concept:

In the power series un = an xn;

limn|un+1un|=limn|(an+1an)|.x

If limn(an+1an)=l, then by ratio test, the series converges, when |lx| is numerically less than 1 i.e., when |x| < 1/l and diverges for other values.

The interval (– 1/l) < x < (1/l) is called the interval of convergence of the power series.

We have to check the convergence at the boundary points also.

Calculation:

Given series is un=x2n(n+1)n 

We have un=x2n(n+1)n and un+1=x2n+2(n+2)n+1 

un+1un=x2n+1n+2nn+1

Now from the ratio test,

limn|un+1un|=limn|x2n+1n+2nn+1|

x2limn|1+1n1+2n11+1n|

⇒ x2;

Therefore the series converges for x2 < 1 -1 < x < 1

Ratio test fails when x2 = 1, therefore at x2 = 1;

At x2 = 1, un=1(n+1)n=1n32.11+1n 

Let’s take vn=1n3/2 

We get limnunvn=limn11+1n=1,afinitequantity 

Therefore, both series will converge or diverge together.

We already know vn is convergent (p-series) ⇒ un will also be convergent;

⇒ at x2 = 1, the series ∑un is convergent;

∴ Interval of convergence is [-1,1] 

Summation Question 2:

Let ∑an be a series and let limnak+1ak=L, then which of the following statements does not express the condition for the ratio test?

  1.  If L < 1, then the series converges
  2. If L > 1, then the series diverges
  3. If L = 0, then the series neither converges nor diverges
  4. If L = 1, then the test fails and is inconclusive

Answer (Detailed Solution Below)

Option 3 : If L = 0, then the series neither converges nor diverges

Summation Question 2 Detailed Solution

Concept:

The Ratio test:

Theorem: Let ∑an be a series and If limnak+1ak=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.

If L = 0 that is L < 1, then the series is convergent.

Summation Question 3:

12n33n4+2 is 

  1. Divergent 
  2. Convergent 
  3. Neither Convergent nor Divergent 
  4. None of these 

Answer (Detailed Solution Below)

Option 1 : Divergent 

Summation Question 3 Detailed Solution

Concept used: 

Suppose  f(x) is positive decreasing continuous function on the interval [1, ∞ ) with f(n) = a

then an  is converging iff 1f(x) dx converges.

Calculations:

1ϕ(x)dx=12x33x4+2dx

=161dtt=16[logt]5=

[Where t = 3x4 + 2]

By integral test, ∑un is divergent.

Summation Question 4:

The given series n=1(1)n1n2n1

  1. Convergent

  2. Divergent
  3. Oscillatory
  4. None of these

Answer (Detailed Solution Below)

Option 3 : Oscillatory

Summation Question 4 Detailed Solution

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) limnun=0

If limnun0, the given series is oscillatory.

Calculation:

Given series is n=1(1)n1n2n1

Now un=n2n1

⇒ unun1=n2n1n12n3=1(2n1)(n3)

So each term is numerically less than its preceding term.

Now limit,

limnun=limnn2n1=12

⇒ limnun0

Series is oscillatory

Summation Question 5:

If S1, S2,.... Sn are the sums of n infinite geometrical series whose first terms are 1, 2, 3, .... n and common ratios are  12,13.14,...,1n+1 , then (S1 + S2 + S3 + ... + Sn) = ?

  1. 12n(n+2)
  2. 13n(n+2)
  3. 12n(n+3)
  4. 13n(n+3)

Answer (Detailed Solution Below)

Option 3 : 12n(n+3)

Summation Question 5 Detailed Solution

Given:
Series 1 = 1+12+14+18+...

Series 2 = 2+2×13+2×132+2×133+...

...

Series n = n+n×1n+1+n×1(n+1)2+n×1(n+1)3+...

Concept:
Sum of an infinite G.P. series, S = a1r , when |r| < 1

Sum of an A.P. series, SAP n(a+l)2 , where, l = last term of series

Calculation:

For Series 1, a = 1, and r = 12
∴ Sa1r = 1112
⇒ S1 = 2

Similarly, for Series 2, a = 2 and r = 13
∴ S1 = 2113
⇒ S2 = 3

Similarly,

S3 = 4

S4 = 5, ...

Sn = 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 + ... + Sn) = n(a+l)2=n(2+n+1)2

∴ S1 + S2 + S3 + ... + Sn ) = n(n+3)2

Summation Question 6:

What is the sum of the first 16 terms of an arithmetic progression if the first term is -9 and last term is 51?

  1. 97
  2. 336
  3. 57
  4. 108

Answer (Detailed Solution Below)

Option 2 : 336

Summation Question 6 Detailed Solution

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

∴ Sum = 336

Summation Question 7:

The series 1+25x+69x2++2n22n+1xn1+(x>0) converges for

  1. 0 < x < 1
  2. 0 < x ≤ 1
  3. x > 1
  4. x ≥ 1

Answer (Detailed Solution Below)

Option 1 : 0 < x < 1

Summation Question 7 Detailed Solution

Concept:

In the power series un = an xn;

limn|un+1un|=limn|(an+1an)|.x

If limn(an+1an)=l, then by ratio test, the series converges, when |lx| is numerically less than 1 i.e., when |x| < 1/l and diverges for other values.

The interval (– 1/l) < x < (1/l) is called the interval of convergence of the power series.

We have to check the convergence at the boundary points also.

Useful concept:

For positive terms series, if limit (n → ∞) un = 0, then it is convergent, if not equal to zero, it is divergent.

Calculation: 

Given series is 1+25x+69x2++2n22n+1xn1+(x>0)

As x > 0, it is a positive term series, so ratio test can be employed.

Now we have

un=2n22n+1xn1,un+1=2n+122n+1+1xn

unun+1=2n22n+1.2n+1+12n+12.xn1xn

unun+1=122n1+12n.2+12n212n.1x

Now

limnunun+1=101+0.2+020.1x=1x

By ratio test, series converges if limit is greater than 1 and diverges if less than 1.

Now the given series converges for 1/x > 1 ⇒ x < 1;

Series diverges for 1/x < 1 ⇒ x > 1;

At x = 1, the test fails,

When x = 1, limnun=limn2n22n+1=limn122n1+12n=10

⇒ ∑un diverges for x = 1

Hence the given series converges for x < 1 and diverges for x ≥ 1;

Summation Question 8:

The given series (221221)1+(332332)2+(443343)3+ is 

  1. Convergent
  2. Divergent
  3. Oscillatory
  4. None of these

Answer (Detailed Solution Below)

Option 1 : Convergent

Summation Question 8 Detailed Solution

Concept:

Cauchy’s Root Test:

In a positive series ∑un, if limn(un)1n=λ

Then the series converges for λ < 1, and diverges for λ > 1; 

Calculation:

Given series is (221221)1+(332332)2+(443343)3+

The general form will be 

∑un ⇒ [(n+1)n+1nn+1n+1n]n

⇒ un=[(n+1)n+1nn+1n+1n]n

Now

un1n=[(n+1)n+1nn+1n+1n]1=(1+1n)1[(1+1n)n1]1

limn(un)1n=limn1.[(1+1n)n1]1

Using the standard limit limn(1+1n)n=e

⇒ limn(un)1n=1e1<1

Since the limit is less than one, the given series is convergent.

Summation Question 9:

For what value of x the following series is convergent.

xx222+x332x442+

  1. The series converges for -1 ≤ x ≤ 1
  2. The series converges for -1 < x ≤ 1
  3. The series diverges for -1 < x < 1
  4. The series converges for x > 1

Answer (Detailed Solution Below)

Option 1 : The series converges for -1 ≤ x ≤ 1

Summation Question 9 Detailed Solution

The given alternating series is:

xx222+x332x442+

un=ft(1n1xnn2

un+1=ft(1)nxn+1ft(n+1)2

ft|un|=ft|x|nn2

ft|un+1|=ft|xn+1ft(n+1)2

ft|un|ft|un+1|=ft|x|nn2ft|x|n+1ft(n+1)2=ft(n+1)2n2×1ft|x|

Now, limnft|un|ft|un+1|=limnft(n+1)2n2×1ft|x|=1ft|x|

Thus the corresponding absolute series is convergent for 1ft|x|>1 i.e. for |x| < 1 i.e. -1 < x < 1

Absolute series is divergent for 1ft|x|<1 i.e. for |x| > 1 i.e. x > 1 and x < -1

Since absolute convergence implies convergence, the given infinite series is convergent for -1 < x < 1

For x = 1, the absolute series becomes

n=1ft|un|=n=11n2

ft|un|=1n2

ft|un+1|=1ft(n+1)2

n < (n+1), for all n

n2<ft(n+1)2

1ft(n+1)2<1n2

ft|un+1|<≤ft|un|; for all n

Moreover, limnft|un|=limn1n2=0

Thus both the conditions of Leibnitz rule re satisfied

So the given series is convergent for x = 1

For x = -1, the absolute series becomes

n=1un=n=11n2

It is evident that the series n=11n2 is convergent as it is a p series with power greater than 1.

Hence the given series is convergent for -1 ≤ x ≤ 1

Summation Question 10:

If (1+ x + x2)nr=02n axr, then a1 − 2a2 + 3a3 − …. −2n a2n is

  1. (n + 1)2n
  2. n
  3. −n
  4. n(2n)

Answer (Detailed Solution Below)

Option 3 : −n

Summation Question 10 Detailed Solution

Concept Used:-

If the summation from 0 to n such that r=0n ar xis given, then the expanded form of it can be written as,

 ⇒ r=0n ar x= a0+a1 x+a2 x2+a3 x3+a4 x4+..........+anxn

Explanation:-

Given,

(1+ x + x2)n = r=02n ar xr

On the expanding right-hand side, we get,

(1+x+x2)n=a0+a1x+a2x2+a3x3+a4x4++a2nx2n

Differentiate it with respect to x,

n(1+x+x2)n1(1+2x)=a1+2a2x+3a3x2+4a4x3++2nanx2x1Now put x = -1 in the above equation,

n(11+1)n1(12)=a12a2+3a34a4+5a52na2nn1(1)n1(1)=a12a2+3a34a4+5a52na2nn=a12a2+3a34a4+5a52na2n

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

Hence, the correct option is 3.

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