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MTH104 ASSIGNMENT NO. 1 SPRING 2023 || 100% RIGHT SOLUTION || CALCULUS AND ANALYTICAL GEOMETRY || BY VuTech

MTH104 ASSIGNMENT NO. 1 SPRING 2023 || 100% RIGHT SOLUTION || CALCULUS AND ANALYTICAL GEOMETRY || BY VuTech

MTH104 ASSIGNMENT NO. 1 SPRING 2023 || 100% RIGHT SOLUTION || CALCULUS AND ANALYTICAL GEOMETRY || BY VuTech

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www.vutechofficial.blogspot.com

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QUESTION-1

By using mathematical induction, prove that the equation   2+4+6+...+2n = n(n+1) is true, for all positive integers. 

SOLUTION:

To prove the equation 2 + 4 + 6 + ... + 2n = n(n + 1) using mathematical induction, we will follow the steps of the proof:


Step 1: Base Case

First, we need to verify that the equation holds true for the smallest possible value of n, which is 1.


For n = 1:

2(1) = 1(1 + 1) -> 2 = 2


Since the equation holds true for n = 1, the base case is satisfied.


Step 2: Inductive Hypothesis

Assume that the equation holds true for some arbitrary positive integer k, where k ≥ 1. This assumption is called the inductive hypothesis.


Assumption: 2 + 4 + 6 + ... + 2k = k(k + 1)


Step 3: Inductive Step

We need to prove that the equation holds true for the next positive integer, which is k + 1.


Consider the sum 2 + 4 + 6 + ... + 2k + 2(k + 1). This can be written as the sum of the first k terms plus the (k + 1)th term.


2 + 4 + 6 + ... + 2k + 2(k + 1) = [2 + 4 + 6 + ... + 2k] + 2(k + 1)


Using the inductive hypothesis, we can substitute k(k + 1) for the sum of the first k terms:


= [k(k + 1)] + 2(k + 1)


Now, let's simplify the right-hand side of the equation:


= k(k + 1) + 2(k + 1)

= k^2 + k + 2k + 2

= k^2 + 3k + 2

= (k + 1)(k + 2)


Therefore, we have shown that if the equation holds true for some positive integer k, then it also holds true for the next positive integer, k + 1.


Step 4: Conclusion

Since the base case (n = 1) holds true and we have shown that if the equation holds true for any positive integer k, it also holds true for k + 1, we can conclude that the equation 2 + 4 + 6 + ... + 2n = n(n + 1) is true for all positive integers.


QUESTION-2

Find the interval satisfying the inequality       
 `4 - 1/2 x >= -7 + 1/4 x`

SOLUTION:

To find the interval satisfying the inequality 4 - (1/2)x ≥ -7 + (1/4)x, we can solve it step by step.


First, let's simplify the equation by combining like terms:


4 - (1/2)x ≥ -7 + (1/4)x


Multiplying both sides of the inequality by 4 to eliminate the fractions:


16 - 2x ≥ -28 + x


Next, let's isolate the variable on one side of the inequality. Subtracting x from both sides:


16 - 2x - x ≥ -28


16 - 3x ≥ -28


Now, let's isolate the variable further by subtracting 16 from both sides:


-3x ≥ -28 - 16


-3x ≥ -44


To solve for x, we need to divide both sides of the inequality by -3. However, since we are dividing by a negative number, the inequality sign will flip:


x ≤ (-44) / (-3)


x ≤ 44/3


The solution to the inequality is x ≤ 44/3. In interval notation, this can be written as (-∞, 44/3] or (-∞, 14.67].




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