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

\(a_r-5a_{r-1}+6a_{r-2}=4^r,\ r\ge2\)

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 \(\mathop {\lim }\limits_{n \to \infty } n\left( {\frac{{{u_n}}}{{{u_{n + 1}}}} - 1} \right) = k\) then

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

The series diverges for k < 1 so is for k < 0 (Option 1)

The series converges for k > 1 (Option 2)

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 -∞.

Concept:

In the power series un = an xn;

\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right| = \mathop {{\rm{lim}}}\limits_{n \to \infty } \left| {\left( {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right)} \right|.x\)

If \(\mathop {{\rm{lim}}}\limits_{n \to \infty } \left( {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right) = l,\) 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:

Here power series is

\(\mathop \sum \nolimits_{n = 1}^\infty \frac{n}{{{{\left( {2n + 1} \right)}^2}}}{\left( {x - 2} \right)^{3n}}\)

= \(\mathop \sum \nolimits_{n = 3}^\infty \frac{{\frac{n}{3}}}{{{{\left( {2\frac{n}{3} + 1} \right)}^2}}}{\left( {x - 2} \right)^n}\)

Now \(\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}}\) will be

= \(\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right)}}{{{{\left[ {\frac{2}{3}\left( {n + 1} \right) + 1} \right]}^2}}}\frac{{{{\left( {\frac{2}{3}n + 1} \right)}^2}}}{n}\)

= \(\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right){\left[ {\frac{{\frac{2}{3}n + 1}}{{\frac{2}{3}\left( {n + 1} \right) + 1}}} \right]^2}\)

= \(\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right){\left[ {\frac{{\frac{2}{3}n + 1}}{{\frac{2}{3}n + \frac{5}{3}}}} \right]^2}\)

= \(\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right){\left[ {\frac{{\frac{2}{3} + \frac{1}{n}}}{{\frac{2}{3} + \frac{5}{3}\frac{1}{n}}}} \right]^2} = 1\)

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

⇒ |x - 2| < R = 1 ⇒ -1 < x -2 < 1

⇒ 1 < x < 3

At x = 3, series becomes

\( \mathop \sum \nolimits_{n = 1}^\infty \frac{n}{{{{\left( {2n + 1} \right)}^2}}}\)

Let u_n = \( \frac{n}{{{{\left( {2n + 1} \right)}^2}}}\)

Let v_n = 1/n;

Now 

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_{n }}}}{{{v_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{n^2}{{{{\left( {2n + 1} \right)}^2}}} = \frac {1}{4}\)

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

\( \mathop \sum \nolimits_{n = 1}^\infty \frac{n}{{{{\left( {2n + 1} \right)}^2}}} (-1)^{3n}= \mathop \sum \nolimits_{n = 1}^\infty \frac{n}{{{{\left( {2n + 1} \right)}^2}}} (-1)^{n}\)

This is an alternating series,

 \(\mathop {\lim }\limits_{n \to \infty } \frac{n}{{{{\left( {2n + 1} \right)}^2}}} = 0 \) 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 \(a_n=ar^n+bnr^n\) 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

Concept:

(i) \(\sum_{x=1}^{n} x= \frac{n(n+1)}{2}\)

(ii) \(\sum_{x=1}^{n} x^2= \frac{n(n+1)(2n+1)}{6}\)

(iii) \(\sum_{x=1}^{n} x^3= (\frac{n(n+1)}{2})^2\)

Explanation:​​​​​​

\(\sum_{x=1}^{n}\)x(x + 2)(x + 4)​​​​

\(\sum_{x=1}^{n}\)x(x²+6x+8) = ​\(\sum_{x=1}^{n}\)x³+6x²+8x​

= ​\(\sum_{x=1}^{n}\)x³+6\(\sum_{x=1}^{n}\)x²+8\(\sum_{x=1}^{n}\)x​

= \((\frac{n(n+1)}{2})^2\)+ 6\(\frac{n(n+1)(2n+1)}{6}\)+8\(\frac{n(n+1)}{2}\)

= n(n+1)\([\frac{n(n+1)}{4}+(2n+1)+4]\)

=n(n+1)\(\frac{n^2+n+8n+4+16}{4}\)

= \(\frac{n(n+1)[n^2+9n+20]}{4}\)

= ​\(\frac{n(n+1)(n+4)(n+5)}{4}\)​

Given :- We have a = 1, l = 11 and \(S_n=36\)

Concept used :- Sum of n term of an A.P. is \(S_n=\dfrac{n}{2}[a+l]\).

Solution :- After using given information, we have

\(36=\dfrac{n}{2}[1+11]=6n\)

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 = \(\mathop {Lt}\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} \)

L > 1 the series is Divergent neither convergent or divergent 

L < 1 the series is Convergent 

L = 1 test fails Neither Convergent nor Divergent 
Calculations:

\(\mathop {Lt}\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} = \mathop {Lt}\limits_{n \to \infty } \left[ {\frac{{{{(n + 1)}^p}}}{{n + 1!}} \times \frac{{n!}}{{{n^p}}}} \right]\, = \mathop {Lt}\limits_{n \to \infty } \left\{ {\frac{1}{{(n + 1)}}{{\left( {\frac{{n + 1}}{n}} \right)}^p}} \right\}\)

\( = \mathop {Lt}\limits_{n \to \infty } \frac{1}{{(n + 1)}} \times \mathop {Lt}\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^p} = 0 < 1;\)

∑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 \(\frac{1}{{1 \cdot 2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + \ldots \infty \) 

Let the sum of n terms be S_n;

The given series can be expanded as

\({S_n} = \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + \ldots + \frac{1}{{n\left( {n + 1} \right)}}\)

\(\Rightarrow {S_n} = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{n} - \frac{1}{{n + 1}}\)

\(\Rightarrow {S_n} = 1 - \frac{1}{{n + 1}} = \frac{n}{{n + 1}}\)

Now limit n tends to infinity, S_n will be

\(\mathop {\lim }\limits_{n \to \infty } \frac{1}{{1 + \frac{1}{n}}} = 1\)