Question

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Prove by using the principle of mathematical induction: – 2 + 5 + 8 + 11 + …+(3n – 1) = 1/2n(3n+1).
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Answer 1

Width of rectangle is 25 cm.

Step-by-step explanation:

Given :-

A wire bend in form of square of side 30 cm.

Then wire is again bend in form of rectangle of length 35 cm.

To find :-

Width of the rectangle.

Solution :-

Here, Concept is : If we are bending wire in form of square than again bending it in rectangle. Than, perimeter of square will equal to perimeter of rectangle because we are not increasing length of wire by one measure we are bending it in square and rectangular shape.

So,

Perimeter of square = 4 × side

⟶ Perimeter = 4 × 30

⟶ Perimeter = 120

Thus,

Perimeter of square is 120 cm.

According to concept, Perimeter of square and perimeter of rectangle are equal.

So, Perimeter of rectangle is 120 cm.

Let, Breadth or width or rectangle be x cm.

We know,

Perimeter of rectangle = 2(Length + Breadth)

⟶ 120 = 2×(35 + x)

⟶ 120 = 70 + 2x

\⟶ 120 - 70 = 2x

⟶ 50 = 2x

⟶ 50/2 = x

⟶ x = 25

We take, Width of rectangle be x.

Therefore,

Width of rectangle is 25 cm.

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Answer 2

Question:⤵️

Prove by using the principle of mathematical induction: – 2 + 5 + 8 + 11 + …+(3n – 1) = 1/2n(3n+1).

Answer:⤵️

Suppose P (n) = 2 + 5 + 8 + 11 + … + (3n – 1) = 1/2 n(3n + 1)

Now let us check for the n = 1,

P (1): 2 = 1/2 × 1 × 4

: 2 = 2

P (n) is true for n = 1.

Then, let us check for the P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k) = 2 + 5 + 8 + 11 + … + (3k – 1) = 1/2 k (3k + 1) … (i)

Therefore,

2 + 5 + 8 + 11 + … + (3k – 1) + (3k + 2)

Then, substituting the value of P (k)

we get, = 1/2 × k (3k + 1) + (3k + 2) by using equation(i)

= [3k2 + k + 2 (3k + 2)] / 2

= [3k2 + k + 6k + 2] / 2

= [3k2 + 7k + 2] / 2

= [3k2 + 4k + 3k + 2] / 2

= [3k (k + 1) + 4(k + 1)] / 2

= [(k + 1) (3k + 4)] /2 P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.

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