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

Associate homogeneous solution is ar - 5ar-1 + 6ar-2 

The characteristic equation of its associated homogeneous relation is

x2 - 5 x + 6 = 0

(x - 3) (x - 2) = 0

∴ x = 3 or x = 2

The solution of associated homogeneous recurrence relationan = 6an-2 - an-1 is

Ah = A1(2)r + A2(3)r

f(r) = 1×4^r it is of the form and 4 is not a root.Therefore it's particular solution is A4r

General solution of recurrence relation is A1(2)r + A2(3)r + A4r

After substitution partial solution in recurrence relation

(A4r) - 5(A4r-1) + 6(A4r-2) = 4r

A - (5A/4) + (6A/16) = 1

∴ A = 8

Therefore general solution is A1(2)r + A2(3)r + 8.4r

Concept:

When the ratio test fails, we apply few tests like Raabe’s test, Logarithmic test and Gauss test.

Raabe’s Test:

In the positive term series ∑ u_n, if  then

The series converges for k > 1 and Diverges for k , but the test fails for k = 1;

The series diverges for k

The series converges for k > 1 (Option 2)

This test fails for k = 1 (Option 3 is wrong)

Concept:

The N^th term test

If , where L is any tangible number other than zero. Then,  diverges.

This is also called the Divergence test.

Calculation:

We have, \)

So, as n → ∞,  → 0

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

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

Concept:

In the power series un = an xn;

If  then by ratio test, the series converges, when |lx| is numerically less than 1 i.e., when |x|

The interval (– 1/l)

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

Calculation:

Here power series is

= 

Now  will be

= 

= 

= 

= 

then by ratio test, the series converges, when |lx| is numerically less than 1

⇒ |x - 2|

⇒ 1

At x = 3, series becomes

Let u_n = 

Let v_n = 1/n;

Now 

The limit is finite, so both the series will have the same behavior.

As v_n is divergent, u_n will also be divergent.

At x = 1, series becomes

This is an alternating series,

  and each term is less than the preceding term, therefore convergent.

Concept:

We use characteristics roots method for repeated roots to solve the above recurrence relation.

If we have an recurrence relation as an + c1an-1 + c2an-2 = 0, then the characteristics equation is given as x2 + c1x + c2 = 0 .

If r is the repeated root of the characteristics equation then the solution to recurrence relation is given as  where a and b are constants determined by initial conditions.

Calculation:

Characteristic equation of recurrence relation a_r = a_r-1 + 2a_r-2 is

x² = x + 2

x² – x – 2 = 0

 (r – 2) (r + 1) = 0

∴ r = 2 or r = -1

The solution of recurrence relation a_r = a_r-1 + 2a_r-2 is

α_n = α₁(2)^r + α₂(-1)^r

Using α₀ = 2

α₀ = α₁(2)⁰ + α₂(-1)⁰ = 2

∴ α₁ + α₂ = 2     ----(1)

Using α₁ = 7

α₁ = α₁(2)¹+ α₂(-1)¹ = 7

∴ 2α₁ - α₂ = 7    ----(2)

By solving (1) and (2)

α₁ = 3 and α₂ = - 1

The solution of recurrence relation a_r = a_r-1 + 2a_r-2 is 3(2)^r – (-1)^r

Concept:

(i) 

(ii) 

(iii) 

Explanation:​​​​​​

x(x + 2)(x + 4)​​​​

x(x²+6x+8) = ​x³+6x²+8x​

= ​x³+6x²+8x​

= + 6+8

= n(n+1)

=n(n+1)

= 

= ​​

Given :- We have a = 1, l = 11 and 

Concept used :- Sum of n term of an A.P. is .

Solution :- After using given information, we have

hence n = 6

Given:

A. M. of 6, 7, 9, x is 10, 

Concept used:

Mean = sum of all observation/total number of observations

Calculation:

A. M. of 6, 7, 9, x is 10, 

⇒ 10 = (6 + 7 + 9 + x)/4

⇒ 40 = 22 + x

∴ x = 18

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 is convergent for all 'p'.

Concept:

Consider the infinite series ∑u_n = u₁ + u₂ + u₃ + … ∞

And let the sum of the first n terms be S_n = u₁ + u₂ + u₃ + … + u_n;

If S_n tends to a finite limit as n → ∞, the series is said to be convergent

If S_n tends to ± ∞ as n → ∞, the series is said to be divergent

If S_n does not tend to a unique limit as n → ∞, the series is said to be oscillatory or non-convergent

Calculation:

Given series is  

Let the sum of n terms be S_n;

The given series can be expanded as

Now limit n tends to infinity, S_n will be

As the S_n tends to a finite limit, the given series is convergent.