Question 1 |

Consider the following recurrence:

\begin{aligned} f(1)&=1; \\ f(2n)&=2f(n)-1, & \text{for }n \geq 1; \\ f(2n+1)&=2f(n)+1, & \text{for }n \geq 1. \end{aligned}

Then, which of the following statements is/are TRUE?

\begin{aligned} f(1)&=1; \\ f(2n)&=2f(n)-1, & \text{for }n \geq 1; \\ f(2n+1)&=2f(n)+1, & \text{for }n \geq 1. \end{aligned}

Then, which of the following statements is/are TRUE?

**MSQ**f(2^n-1)=2^n-1 | |

f(2^n)=1 | |

f(5 \dot 2^n)=2^{n+1}+1 | |

f(2^n+1)=2^n+1 |

Question 1 Explanation:

Question 2 |

Consider the recurrence relation a_{1}=8, a_{n}=6n^{2}+2n+a_{n-1}. Let a_{99}= K \times 10^{4}. The value of K is .

198 | |

148 | |

226 | |

312 |

Question 2 Explanation:

Question 3 |

Let a_{n} be the number of n-bit strings that do NOT contain two consecutive 1s. Which one of the following is the recurrence relation for a_{n}?

a_{n}=a_{n-1}+2a_{n-2} | |

a_{n}=a_{n-1}+a_{n-2} | |

a_{n}=2a_{n-1}+a_{n-2} | |

a_{n}=2a_{n-1}+2a_{n-2} |

Question 3 Explanation:

Question 4 |

Let a_{n} represent the number of bit strings of length n containing two consecutive 1s. What is the recurrence relation for a_{n}?

a_{n-2}+a_{n-1}+2^{n-2} | |

a_{n-2}+2a_{n-1}+2^{n-2} | |

2a_{n-2}+a_{n-1}+2^{n-2} | |

2a_{n-2}+2a_{n-1}+2^{n-2} |

Question 4 Explanation:

Question 5 |

Consider the sequence \left \langle x_n \right \rangle,\; n \geq 0 defined by the recurrence relation x_{n + 1} = c \cdot (x_n)^2 - 2, where c > 0.

For which of the following values of c, does there exist a non-empty open interval (a, b) such that the sequence x_n converges for all x_0 satisfying a < x_0 < b?

i. 0.25

ii. 0.35

iii. 0.45

iv. 0.5

For which of the following values of c, does there exist a non-empty open interval (a, b) such that the sequence x_n converges for all x_0 satisfying a < x_0 < b?

i. 0.25

ii. 0.35

iii. 0.45

iv. 0.5

i only | |

i and ii only | |

i, ii and iii only | |

i, ii, iii and iv |

Question 5 Explanation:

Question 6 |

Consider the sequence \langle x_n \rangle , \: n \geq 0 defined by the recurrence relation x_{n+1} = c . x^2_n -2, where c > 0.

Suppose there exists a non-empty, open interval (a, b) such that for all x_0 satisfying a \lt x_0 \lt b, the sequence converges to a limit. The sequence converges to the value?

Suppose there exists a non-empty, open interval (a, b) such that for all x_0 satisfying a \lt x_0 \lt b, the sequence converges to a limit. The sequence converges to the value?

\frac{1+\sqrt{1+8c}}{2c} | |

\frac{1-\sqrt{1+8c}}{2c} | |

2 | |

\frac{2}{2c-1} |

Question 6 Explanation:

Question 7 |

Let H_1, H_2, H_3, ... be harmonic numbers. Then, for n \in Z^+, \sum_{j=1}^{n} H_j can be expressed as

nH_{n+1} - (n + 1) | |

(n + 1)H_n - n | |

nH_n - n | |

(n + 1) H_{n+1} - (n + 1) |

Question 7 Explanation:

There are 7 questions to complete.