Sequences and Series MCQ Quiz in తెలుగు - Objective Question with Answer for Sequences and Series - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Given:

Concept used: 
The necessary condition of series to be convergent is ;

Limit comparison test:

if un and vn are positive terms sequence such that ">Ltn→∞unvn=c" id="MathJax-Element-6-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">Ltn→∞unvn=cLtn→∞unvn=c

 where c > 0 and c is finite then Both un and vn converge and diverge together

P- series test: 

∑ ">1np" id="MathJax-Element-25-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">1np1np is convergent for p > 1 and divergent for p ≤  1

u_n =  

⇒ necessary condition fails 

Now, by limit comparison test 

Take 

 (Finite)

∑v_n is divergent by p-series test. (p = 0

∴ By limit comparison test, ∑u_n is divergent

∴ option 2 is correct 

Concept:

Calculation:

We know that,

Therefore,

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  ar xr is given, then the expanded form of it can be written as,

 ⇒  ar xr = a₀+a₁ x+a₂ x²+a₃ x³+a₄ x⁴+..........+a_nxⁿ

Explanation:-

Given,

(1+ x + x2)n =  ar xr, 

On the expanding right-hand side, we get,

Differentiate it with respect to x,

Now put x = -1 in the above equation,

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

Hence, the correct option is 3.

Given:

The Sequence mn where m, n ∈ N 

Concept used:

The Sequence is bounded if for all the Terms of the sequence there exists M and m such that m i 

if no such m and M exists then Sequence is Unbounded 

if revolving around a number than  it is Oscillating.

Calculations:

The Terms of the Sequence mⁿ are as 0, 1, 2, 4, 8,.......,3, 9, 27,........, 4, 16, 64,........ as so on.

So there doesn't exist M such that all the Terms of the Sequence are less than that M and m = 0 here'

∴ The sequence is not bounded i.e Unbounded.

Concept used:

Ratio test:

L = 

P - Series test: 

∴ By ratio test, ∑un converges when x 1.

When x = 1,  

Take  ; By Limit comparison test ∑u_n is convergent 

∴ ∑u_n is convergent if x ≤  1 and divergent if x > 1.

Given:

Concept used:

Limit Comparision test:

if an and bn are two positive series such that ">Ltn→∞anbn=c" id="MathJax-Element-24-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0"> 

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

P - series test:

 ∑ is convergent for p > 1 and divergent for p ≤  1

Calculations:

take v_{n =} 

 (where t = 1/n) = 1

∴ ∑u_n, ∑v_n both converge or diverge. But ∑v_n =  is divergent

(p-series test, p = 1);

∴ ∑u_n is divergent.

Concept:

By ratio test, if 

Then the series converges for λ > 1 and diverges for λ

By Raabe’s test, if 

Then the series converges for k > 1 and diverges for k

Calculation:

From the given series,

Taking limit,

Thus by ratio test, series converges for x² 2 > 4, but fails for x² = 4;

When x² = 4,

Thus by Raabe's test, the series diverges.

Hence the given series

converges for x²

diverges for x² ≥ 4 → x ≤ -2 and x ≥ 2

Concept used:

Ratio test:

L = 

L > 1 the series is Divergent neither convergent or divergent 

L

L = 1 test fails Neither Convergent nor Divergent 

Calculations:

u_n = n^1-n ; u_{n + 1} = (n + 1)^-n ;

∴ By ratio test ∑u_n, is convergent

Given:

Concept used:

Limit Comparision test:

if an and bn are two positive series such that  

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

P - Series test:

∑ is convergent for p > 1 and divergent for p ≤  1

Calculations:

Let v_n  =  , so that Σv_n is convergent by p-series test.

where t = 1/n, Thus 

∴ By Limit comparison test Σu_n is convergent.