मराठी

1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) = N ( 4 N 2 + 6 N − 1 ) 3 - Mathematics

Advertisements
Advertisements

प्रश्न

1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]

 
Advertisements

उत्तर

Let P(n) be the given statement.
Now,

\[P(n) = 1 . 3 + 3 . 5 + 5 . 7 + . . . + (2n - 1)(2n + 1) = \frac{n(4 n^2 + 6n - 1)}{3}\]

\[\text{ Step }  1: \]

\[P(1) = 1 . 3 = 3 = \frac{1(4 \times \left( 1 \right)^2 + 6 \times 1 - 1)}{3}\]

\[\text{ Hence, P(1) is true }  . \]

\[\text{ Step 2: }  \]

\[\text{ Let P(m) be true} . \]

\[\text{ Then,}  \]

\[1 . 3 + 3 . 5 + . . . + (2m - 1)(2m + 1) = \frac{m(4 m^2 + 6m - 1)}{3}\]

\[\text{ To prove: P(m + 1) is true}  . \]

\[\text{ That is, }  \]

\[1 . 3 + 3 . 5 + . . . + (2m + 1)(2m + 3) = \frac{(m + 1)\left[ 4(m + 1 )^2 + 6\left( m + 1 \right) - 1 \right]}{3}\]

\[ \text{ Now, P(m) is equal to: }  \]

\[1 . 3 + 3 . 5 + . . . + (2m - 1)(2m + 1) = \frac{m(4 m^2 + 6m - 1)}{3}\]

\[ \Rightarrow 1 . 3 + 3 . 5 + . . . + (2m - 1)(2m + 1) + (2m + 1)(2m + 3) = \frac{m(4 m^2 + 6m - 1)}{3} + (2m + 1)(2m + 3) \left[ \text{ Adding } (2m + 1)(2m + 3) \text{ to both sides }  \right]\]

\[ \Rightarrow P(m + 1) = \frac{m(4 m^2 + 6m - 1) + 3(4 m^2 + 8m + 3)}{3}\]

\[ \Rightarrow P(m + 1) = \frac{4 m^3 + 6 m^2 - m + 12 m^2 + 24m + 9}{3} = \frac{4 m^3 + 18 m^2 + 23m + 9}{3}\]

\[ \Rightarrow P(m + 1) = \frac{4m( m^2 + 2m + 1) + 10 m^2 + 19m + 9}{3}\]

\[ = \frac{4m(m + 1 )^2 + (10m + 9)(m + 1)}{3}\]

\[ = \frac{(m + 1)\left[ 4m(m + 1) + 10m + 9 \right]}{3}\]

\[ = \frac{(m + 1)}{3}(4 m^2 + 8m + 4 + 6m + 5)\]

\[ = \frac{(m + 1)\left[ 4(m + 1 )^2 + 6\left( m + 1 \right) - 1 \right]}{3}\]

\[\text{ Thus, P(m + 1) is true .}  \]

\[\text{ By the principle of mathematical induction, P(n) is true for all n}  \in N .\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 12: Mathematical Induction - Exercise 12.2 [पृष्ठ २७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 12 Mathematical Induction
Exercise 12.2 | Q 13 | पृष्ठ २७

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Prove the following by using the principle of mathematical induction for all n ∈ N

`1 + 3 + 3^2 + ... + 3^(n – 1) =((3^n -1))/2`


Prove the following by using the principle of mathematical induction for all n ∈ N

`1+ 1/((1+2)) + 1/((1+2+3)) +...+ 1/((1+2+3+...n)) = (2n)/(n +1)`

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2)  = `(n(n+1)(n+2)(n+3))/(4(n+3))`


Prove the following by using the principle of mathematical induction for all n ∈ N

1.3 + 2.3^3 + 3.3^3  +...+ n.3^n = `((2n -1)3^(n+1) + 3)/4`

Prove the following by using the principle of mathematical induction for all n ∈ N

1.3 + 3.5 + 5.7 + ...+(2n -1)(2n + 1) = `(n(4n^2 + 6n -1))/3`

Prove the following by using the principle of mathematical induction for all n ∈ N

`1/2.5 + 1/5.8 + 1/8.11 + ... + 1/((3n - 1)(3n + 2)) = n/(6n + 4)`

Prove the following by using the principle of mathematical induction for all n ∈ N

1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ...+ `1/(n(n+1)(n+2)) = (n(n+3))/(4(n+1) (n+2))`

Prove the following by using the principle of mathematical induction for all n ∈ N: `1+2+ 3+...+n<1/8(2n +1)^2`


Prove the following by using the principle of mathematical induction for all n ∈ Nx2n – y2n is divisible by x y.


If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.

 

12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]

 

32n+7 is divisible by 8 for all n ∈ N.

 

52n+2 −24n −25 is divisible by 576 for all n ∈ N.

 

11n+2 + 122n+1 is divisible by 133 for all n ∈ N.

 

\[\frac{1}{2}\tan\left( \frac{x}{2} \right) + \frac{1}{4}\tan\left( \frac{x}{4} \right) + . . . + \frac{1}{2^n}\tan\left( \frac{x}{2^n} \right) = \frac{1}{2^n}\cot\left( \frac{x}{2^n} \right) - \cot x\] for all n ∈ and  \[0 < x < \frac{\pi}{2}\]

 


\[1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + . . . + \frac{1}{n^2} < 2 - \frac{1}{n}\] for all n ≥ 2, n ∈ 

 


\[\sin x + \sin 3x + . . . + \sin (2n - 1)x = \frac{\sin^2 nx}{\sin x}\]

 


Prove that the number of subsets of a set containing n distinct elements is 2n, for all n \[\in\] N .

 

\[\text{ A sequence }  a_1 , a_2 , a_3 , . . . \text{ is defined by letting }  a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]


\[\text{ A sequence } x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_0 = 5 and x_k = 4 + x_{k - 1}\text{  for all natural number k . } \]
\[\text{ Show that } x_n = 5 + 4n \text{ for all n }  \in N \text{ using mathematical induction .} \]


Prove by method of induction, for all n ∈ N:

12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`


Prove by method of induction, for all n ∈ N:

13 + 33 + 53 + .... to n terms = n2(2n2 − 1)


Prove by method of induction, for all n ∈ N:

(24n−1) is divisible by 15


Prove by method of induction, for all n ∈ N:

3n − 2n − 1 is divisible by 4


Prove by method of induction, for all n ∈ N:

(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)


Answer the following:

Prove, by method of induction, for all n ∈ N

2 + 3.2 + 4.22 + ... + (n + 1)2n–1 = n.2n 


Answer the following:

Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2


Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:

`(1 - 1/2^2).(1 - 1/3^2)...(1 - 1/n^2) = (n + 1)/(2n)`, for all natural numbers, n ≥ 2. 


Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:

2n + 1 < 2n, for all natual numbers n ≥ 3.


Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is ______.


Give an example of a statement P(n) which is true for all n. Justify your answer. 


Prove the statement by using the Principle of Mathematical Induction:

n3 – 7n + 3 is divisible by 3, for all natural numbers n.


Prove the statement by using the Principle of Mathematical Induction:

32n – 1 is divisible by 8, for all natural numbers n.


Prove the statement by using the Principle of Mathematical Induction:

2 + 4 + 6 + ... + 2n = n2 + n for all natural numbers n.


Prove that for all n ∈ N.
cos α + cos(α + β) + cos(α + 2β) + ... + cos(α + (n – 1)β) = `(cos(alpha + ((n - 1)/2)beta)sin((nbeta)/2))/(sin  beta/2)`.


Prove that `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24`, for all natural numbers n > 1.


Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.


For all n ∈ N, 3.52n+1 + 23n+1 is divisible by ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×