MATHS M.C.Q (1 Marks)

1. 5

Solution:

Given,

$10\text{n }+3\times^{\text{n}+2}+\lambda$ is divisible by 9,

$\text{P}(1)=10^1+3\times4^{1+2}+\lambda$ is exactly divisible by 9 then the value of $\lambda$ is 5.

4. 0

Solution:

$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4+...\text{i}^{1000}$

$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4 \ [\because\text{i}^2=-1,\text{i}^3=-\text{i and} \ \text{i}^4=1]$

$=\text{i}-1-\text{i}+1$

$=0$

Similarly, the sum of the next four terms of the series will be equal to 0. This is because the powers of i follow a cyclicity of 4. Hence, the sum of all terms, till 1000, will be zero.

$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4+...\text{i}^{1000}=0$

3. For all $\text{n}\geq5$

Solution:

P(n) is true for all positive integer n, i.e. $\text{n}\geq5,$

Where P(n) is a Propositional function, complete two steps:

Basic Step: Verify that the proposition P(1) is true.

Inductive Step: Show the conditional statement,

$\big[\text{P}(1) \wedge \text{P}(2) \wedge-\wedge\text{P}(\text{k})\big]\rightarrow\text{P(k+1)}$ holds for all positive integer.

1. 1

Solution:

Given

xⁿ - 1

We know that

x = k is the root of the equation(x - 1)

⇒ xⁿ - 1 = 0

⇒ xⁿ = 1

Hence, the least positive integral value of $\lambda$ is 1.

2. 17

Solution:

3.5^{2n+ 1} + 2³ⁿ⁺¹ is divisible by 17, $\text{n}\in\text{N}$

Step 1: 3.5²⁽¹⁾⁺¹ + 2^{3(1) + 1}

3.5³ + 2⁴ = 391

Step 2: Assuming True for n = k

Hence, it is proved that 3.5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17.

3. -1

Solution:

(49)ⁿ + 16n - 1

⇒ (1 + 48)ⁿ + 16n - 1

⇒ 1 + 48n + ... 48ⁿ + 16n - 1

⇒ 64n + nC₂(48)² + nC₃(48)³ + ... + (48)ⁿ

⇒ 64(n + nC₂(6)² + nC₃(6)³48 + ... + (6)ⁿ 8^n-2)

$\therefore$ 49ⁿ + 16n - 1 is divisible by 64