18. <math><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>3</mn><mo>+</mo><mo>?</mo><mo>+</mo><mi>n</mi><mo><</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>8</mn></mrow></mfrac><msup><mrow><mo stretchy="false">(</mo><mn>2</mn><mi>n</mi><mo>+</mo><mn>1</mn><mo stretchy="false">)</mo></mrow><mrow><mn>2</mn></mrow></msup></mrow></math>

18. We can write the given statement asp(n):1+2+3+?+n<18(2n+1)2If n = 1, we get, P(1): 1 < 18 (2k + 1)2= 1< 18 (3)2= 1 < 98Which is true.Consider P(k) be true some positive integer k1+ 2 + …. + k< 18 (2k + 1)2 (1)Let us prove P(k +1) is true.Here,(1 + 2 +…. k)+ (k +1) < 18 (2k + 1)2+ (k +1)By using (1),<18{(2k+1)2+8(k+1)}<18{(2k)2+2⋅2k⋅+12+8k+8}<18{4k2+4k+1+8k+8}<18{4k2+12k+9}So, we get,< 18 {2k+ 3}2< 18 {2(k +1) +1}2(1 + 2 + 3 + … + k) + (k + 1) < 18 (2k +1)2+ (k +1)P(k + 1) is true whenever P(k) is true.∴ Hence, from the Principle of mathematical induction, the P(k) is true for all natural number n.

Ask & Answer Question

Ncert Solutions Maths class 11th Maths Class 11th Principle of Mathematical Induction

18. $1 + 2 + 3 + ? + n < \frac{1}{8} {(2 n + 1)}^{2}$

2 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

4 months ago

18. We can write the given statement as

$p (n) : 1 + 2 + 3 + ? + n < \frac{1}{8} {(2 n + 1)}^{2}$

If n = 1, we get,

P(1): 1 < $\frac{1}{8}$ (2k + 1)²= 1< $\frac{1}{8}$ (3)²

= 1 < $\frac{9}{8}$

Which is true.

Consider P(k) be true some positive integer k

1+ 2 + …. + k< $\frac{1}{8}$ (2k + 1)²(1)

Let us prove P(k +1) is true.

Here,

(1 + 2 +…. k)+ (k +1) < $\frac{1}{8}$ (2k + 1)²+ (k +1)

By using (1),

$< \frac{1}{8} {{(2 k + 1)}^{2} + 8 (k + 1)}$

$< \frac{1}{8} {{(2 k)}^{2} + 2 \cdot 2 k \cdot + 1^{2} + 8 k + 8}$

$< \frac{1}{8} {4 k^{2} + 4 k + 1 + 8 k + 8}$

$< \frac{1}{8} {4 k^{2} + 12 k + 9}$

So, we get,

< $\frac{1}{8}$ {2k+ 3}²

< $\frac{1}{8}$ {2(k +1) +1}²

(1 + 2 + 3 + … + k) + (k + 1)

...more

18. We can write the given statement as

$p (n) : 1 + 2 + 3 + ? + n < \frac{1}{8} {(2 n + 1)}^{2}$

If n = 1, we get,

P(1): 1 < $\frac{1}{8}$ (2k + 1)²= 1< $\frac{1}{8}$ (3)²

= 1 < $\frac{9}{8}$

Which is true.

Consider P(k) be true some positive integer k

1+ 2 + …. + k< $\frac{1}{8}$ (2k + 1)²(1)

Let us prove P(k +1) is true.

Here,

(1 + 2 +…. k)+ (k +1) < $\frac{1}{8}$ (2k + 1)²+ (k +1)

By using (1),

$< \frac{1}{8} {{(2 k + 1)}^{2} + 8 (k + 1)}$

$< \frac{1}{8} {{(2 k)}^{2} + 2 \cdot 2 k \cdot + 1^{2} + 8 k + 8}$

$< \frac{1}{8} {4 k^{2} + 4 k + 1 + 8 k + 8}$

$< \frac{1}{8} {4 k^{2} + 12 k + 9}$

So, we get,

< $\frac{1}{8}$ {2k+ 3}²

< $\frac{1}{8}$ {2(k +1) +1}²

(1 + 2 + 3 + … + k) + (k + 1) < $\frac{1}{8}$ (2k +1)²+ (k +1)

P(k + 1) is true whenever P(k) is true.

$∴$ Hence, from the Principle of mathematical induction, the P(k) is true for all natural number n.

less

0 Upvotes 0 Downvotes

Similar Questions for you

From the base of a pole of height 20 meter, the angle of elevation of the top of tower is 60°. The pole subtends an angle 30° at the top of the tower. Then the height of the tower is:

Vishal Baghel

Let base = b

$t a n 60 ° = \frac{h}{b}$

$t a n 30 ° = \frac{h - 20}{b}$

If $\underset{x \to 1}{l i m} \frac{s i n (3 x^{2} - 4 x + 1) - x^{2} + 1}{2 x^{3} - 7 x^{2} + a x + b} = - 2,$ then the value of (a – b) is equal to………….

alok kumar singh

$\underset{x \to 1}{l i m} \frac{s i n (3 x^{2} - 4 x + 1) - x^{2} + 1}{2 x^{3} - 7 x^{2} + a x + b} = - 2$

For this limit to be defined 2x³ – 7x² + ax + b should also trend to 0 or x ® 1.

$∴ 2 . {(1)}^{3} - 7 {(1)}^{2} + a \cdot 1 + b = 0$

2 – 7 + (a + b) = 0

(a + b) = 5 …………….(i)

Now this becomes % form $∴$ we apply L’lopital rule

$\underset{x \to 1}{l i m} \frac{(3 x^{2} - 4 x + 1) - x^{2} + 1}{2 x^{3} - 7 x^{2} + a x + b} = \underset{x \to 1}{l i m} \frac{c o s (3 x^{2} - 4 x + 1) (6 x - 4) - 2 x}{6 x^{2} - 14 x + a}$

Now the numerator again ® 0 as x = 1

$∴$ 6x² – 14x + a ® 0 as x = 1

6 . (1)² – 14 + a = 0

a = 8 …………….(ii)

a + b = 5 $∴ a - b = 8 - (- 3) = 11$

(b = -3) ® from (i) & (ii)

Let the solution curve y = y(x) of the differential equation (4 + x²)dy – 2x(x² + 3y + 4)dx = 0 pass through the origin. They y(2) is equal to

Payal Gupta

$(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$

$\Rightarrow \frac{d y}{d x} = (\frac{6 x}{x^{2} + 4}) y + 2 x$

$e^{- 3 l n (x^{2} + 4)} = \frac{1}{{(x^{2} + 4)}^{3}}$

$\frac{y}{{(x^{2} + 4)}^{3}} = \int \frac{2 x}{{(x^{2} + 4)}^{3}} d x + c$

$\Rightarrow y = - \frac{1}{2} (x^{2} + 4) + c {(x^{2} + 4)}^{3}$

When x = 0, y = 0 gives

$c = \frac{1}{32}$

So, for x = 2, y = 12

Let $X = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}], Y = α I + β X + γ X^{2} a n d$ $Z = α^{2} I - α β X + (β^{2} - α γ) X^{2}, α, β, γ \in R .$ $^{}^{}^{}$ If $^{} \begin{matrix} \frac{}{} & \frac{}{} & \frac{}{} \end{matrix}$ $Y^{- 1} = [\begin{matrix} \frac{1}{5} & \frac{- 2}{5} & \frac{1}{5} \\ 0 & \frac{1}{5} & \frac{- 2}{5} \\ 0 & 0 & \frac{1}{5} \end{matrix}]$ , then ${(α - β + γ)}^{2}$ $^{}$ is equal to………..

Vishal Baghel

$? X = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}]$

$∴ X^{2} = [\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

$∴ Y = α l + β X + γ X^{2} = [\begin{matrix} α & β & γ \\ 0 & α & β \\ 0 & 0 & α \end{matrix}]$

$\begin{array}{l} ∴ α = 5, β = 10, y = 15 \\ ∴ {(α - β + y)}^{2} = 100 \end{array}$

24. 2n+7<(n 3)²

Payal Gupta

24. Let P (n) be the statement “ 2n+7< (n+3)2”

ofn=1

P (1): 2 $\times 1 + 7 < {(1 + 3)}^{2}$

$9 < 4^{2}$

9<16 which is true. This P (1) is true.

Suppose P (k) is true.

P (k)= 2k+7< (k+3)²^{. (1)}

Lets prove that P (k +1) is also true.

“ 2 (k + 1) + 7 < (k + 4)²=k²+ 8k + 16”

P (k +1) = 2 (k +1) +7 = (2k +7) +2

< (k +3)²+ 2 (Using 1)

= k²+ 9 + 6k +2 = k²+6k +11

Adding and subtracting (2k + k) in the R. H. S.

$= k^{2} + 6 k + 11 + 2 k + 5 - (2 k - 5)$

$= (k^{2} - 8 k + 16) - (2 k - 5)$

$= {(k + 4)}^{2} - (2 k - 5)$

$< {(k + 4)}^{2}, since 2 k + 5 > 0$ for all $k \in N$

$\Rightarrow P (k + 1)$ is true.

$∴$ By the principle of mathematical induction, P (n)is true for all n $\in$ N.

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

18. $1 + 2 + 3 + ? + n < \frac{1}{8} {(2 n + 1)}^{2}$

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

18. 1+2+3+?+n<18(2n+1)2

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

18. $1 + 2 + 3 + ? + n < \frac{1}{8} {(2 n + 1)}^{2}$