Generating Functions MCQ Quiz in తెలుగు - Objective Question with Answer for Generating Functions - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

a_n = a_n-1 + 2n (given)

a₀ = 2, a₁ = 2 + 2(1) = 4

a₂ = 4 + 4 = 8, a₃ = 8 + 6 = 14

a₄ = 14 + 8 = 22, a₅ = 22 + 10 = 32

∴ sequence: 2, 4, 8, 14, 22, 32…

G(x) = 2 + 4x + 8x² + 14x³ + 22x⁴ + 32x⁵…

xG(x) = 2x + 4x² + 8x³ + 14x⁴ + 22x⁵ …

G(x) – xG(x) = 2 + 2x + 4x² + 6x³ + 8x⁴ + 10x⁵…

G(x)(1 – x ) = 2 + 2x( 1 + 2x + 3x² + 4x³ …)

$G\left(x\right)\left(1-x\right)=2+2x\times \frac{1}{{\left(1-x\right)}^2}$

$G\left(x\right)=\frac{2}{1-x}+\frac{2x}{{\left(1-x\right)}^3}$

x₁ + x₂ + x₃ + x₄ = 7

∴ n = 4 and r = 7

${(n+r-1)}_{C_r}$

a_n = 3n + 5 (given)

a₀ = 3(0) + 5 = 5, a₁ = 3(1) + 5 = 8

a₂ = 3(2) + 5 = 11, a₃ = 3(3) + 5 = 14

a₄ = 3(4) + 5 = 17, a₅ = 3(5) + 5 = 20

∴ sequence: 5, 8, 11, 14, 17, 20…

G(x) = 5 + 8x + 11x² + 14x³ + 17x⁴ + 20x⁵…

xG(x) = 5x + 8x² + 11x³ + 14x⁴ + 17x⁵ …

G(x) – xG(x) = 5 + 3x + 3x² + 3x³ …

G(x)(1 – x ) = 5 + 3x( 1 + x + x² + x³ …)

$G\left(x\right)\left(1-x\right)=5+3x\times \left(\frac{1}{1-x}\right)$

$G\left(x\right)=\frac{5}{1-x}+3x\times \frac{1}{{\left(1-x\right)}^2}$

$G\left(x\right)=\frac{5-5x+3x}{{\left(1-x\right)}^2}$

$G\left(x\right)=\frac{5-2x}{{\left(1-x\right)}^2}$

Solution of recurrence relation: a_n = (c₁ + c₂n + c₃n²) (-2)ⁿ

∴ characteristic equation of recurrence relation: (r + 2)³ = 0

By solving the characteristic equation of recurrence relation

(r + 2)³ = 0

∴ r³ + 6r² + 12r + 8 = 0                      

r³ = -(6r² + 12r + 8)

a_n = -2(3a_n-1 + 6a_n-2 + 4_n-3)

We can construct a bit string of length eight that begins with a 1 in 2⁷ =128 ways.

Similarly, we can construct a bit string of length eight ending with the two bits 00, in 2⁶ =64 ways. 

Some of the ways to construct a bit string of length eight starting with a 1 are the same as the ways to construct a bit string of length eight that ends with the two bits 00. There are 2⁵ =32 ways to construct such a string.

The number of ways to construct a bit string of length eight that begins with a 1 ort hat ends with 00 = 128+64−32=160. 

a_n = a_{n – 1} + 6n² + 2n

= a_n-2 + 6(n-1)² + 2(n-1) + 6n² + 2n

$\begin{array}{l}\equiv a_1+6\sum _{i=2}^n{i}^2+2\sum _{i=2}^{n}i\\ \equiv 8+6\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}-1\right]+2\left[\frac{n\left(n+1\right)}{2}-1\right]\\ \equiv 8+6\left[\frac{n\left(n+1\right)\left(2n+1\right)-6}{6}\right]+2\left[\frac{n\left(n+1\right)-2}{2}\right]\end{array}$

≡ 8 + n (n+1) (2n+1) – 6 + n(n+1) – 2

a_n = n(n + 1) (2n + 2)

a_aa = k × 10⁴

Also,

a_aa = 99 (99 + 1) (198 + 2)

= 99 × 10² × 2 × 10²

= 198 × 10⁴

Prime factorization of 2100 is: 2². 3¹. 5². 7¹

∴ # divisors = (2 + 1) (1 + 1) (2 + 1) (1 + 1)

 = 3 × 2 × 3 × 2

 = 36

Minimum divisor is 1 and Maximum divisor is 2100.

Concept:

Every bounded infinite set of real numbers has at least one limit point or cluster point. The theorem is known as Bolzano-Weierstrass Theorem

Some other important theorems by Bolzano-Weierstrass are:

1. Every bounded sequence has a convergent subsequence.

2. A sequence {an} converges if and only if it is unbounded and has exactly one subsequential limit

Additional Information

Cauchy's theorem

If f(z) is analytic everywhere within a simply-connected region then:

${\oint }_{c}f(z)dz=0$

for every simple closed path C lying in the region

Although base condition is not mentioned for tower of Hanoi when number disc is 0, that is, n = 0

number of moves needed is 0.

T(0) = 0 (base condition for Tower of Hanoi)

T(1) = 2 × T(0) + 1 = 0 + 1 = 1

T(2) = 2 × T(1) + 1 = 2 + 1 = 3

T(3) = 2 × T(2) + 1 = 6 + 1 = 7

T(4) = 2 × T(3) + 1 = 14 + 1 = 15

T(5) = 2 × T(4) + 1 = 30 + 1 = 31

T(6) = 2 × T(5) + 1 = 62 + 1 = 63

T(7) = 2 × T(6) + 1 = 126 + 1 = 127

Tips and Tricks:

Number of moves required for n disc in a Tower of Hanoi is 2ⁿ – 1 = 2⁷ – 1 = 127.

The characteristic equation of  $a_n=8{a_n}_{-1}-16{a_n}_{-2}$ is

$r^2-8r+16=0$

$\left(r-4\right)\left(r-4\right)=0$

$a_n={\alpha }_1{\left(r_1\right)}^n+{\alpha }_{2}n{\left(r_1\right)}^n$

$a_n={\alpha }_1{\left(4\right)}^n+{\alpha }_{2}n{\left(4\right)}^n$