हिंदी

Prove by method of induction, for all n ∈ N: 5 + 52 + 53 + .... + 5n = 54(5n-1)

Advertisements
Advertisements

प्रश्न

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

5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`

योग
Advertisements

उत्तर

Let P(n) ≡ 5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`, for all n ∈ N

Step I:

Put n = 1

L.H.S. = 5

R.H.S. = `5/4(5^1 - 1)` = 5 = L.H.S.

∴ P(n) is true for n = 1.

Step II:

Let us consider that P(n) is true for n = k.

∴ 5 + 52 + 53 + …. + 5k = `5/4(5^"k" - 1)`   ...(i)

Step III:

We have to prove that P(n) is true for n = k + 1 i.e., to prove that

5 + 52 + 53 + …. + 5k+1 = `5/4(5^("k"+1) - 1)`

L.H.S. = 5 + 52 + 53 + …. + 5k+1 

= 5 + 52 + 53 + …. + 5k + 5k+1 

= `5/4(5^"k" - 1) + 5^("k"+1)`   ...[From (i)]

= `(5.5^"k" - 5 + 4.5^("k"+1))/4`

= `(5^("k"+1) + 4.5^("k"+1) - 5)/4`

= `(5.5^("k"+1) - 5)/4`

= `5/4(5^("k" + 1) - 1)`

= R.H.S.

∴ P(n) is true for n = k + 1

Step IV:

From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ 5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`, for all n ∈ N.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Methods of Induction and Binomial Theorem - Exercise 4.1 [पृष्ठ ७४]

APPEARS IN

बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
अध्याय 4 Methods of Induction and Binomial Theorem
Exercise 4.1 | Q 13 | पृष्ठ ७४

संबंधित प्रश्न

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

1.2 + 2.3 + 3.4+ ... + n(n+1) = `[(n(n+1)(n+2))/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

`a + ar + ar^2 + ... + ar^(n -1) = (a(r^n - 1))/(r -1)`

Prove the following by using the principle of mathematical induction for all n ∈ N (2+7) < (n + 3)2


If P (n) is the statement "n(n + 1) is even", then what is P(3)?


1 + 2 + 3 + ... + n =  \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .


2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]

 

1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]

 

\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]


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

 

a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]

 


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

 

Given \[a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and }  a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right)\] for n ≥ 2, where a > 0, A > 0.
Prove that \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]

 

Prove that n3 - 7+ 3 is divisible by 3 for all n \[\in\] N .

  

7 + 77 + 777 + ... + 777 \[{. . . . . . . . . . .}_{n - \text{ digits } } 7 = \frac{7}{81}( {10}^{n + 1} - 9n - 10)\]

 

\[\frac{n^{11}}{11} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{62}{165}n\] is a positive integer 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  \[0 < x < \frac{\pi}{2}\]

 


Let P(n) be the statement : 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all n ∈ N


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

 


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

 

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:

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


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

`[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]` ∀ n ∈ N


Answer the following:

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

8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`


Answer the following:

Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.


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

22n – 1 is divisible by 3.


Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer


Prove the statement by using the Principle of Mathematical Induction:

23n – 1 is divisible by 7, for all natural numbers n.


Prove the statement by using the Principle of Mathematical Induction:

For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.


Prove the statement by using the Principle of Mathematical Induction:

1 + 5 + 9 + ... + (4n – 3) = n(2n – 1) for all natural numbers n.


A sequence a1, a2, a3 ... is defined by letting a1 = 3 and ak = 7ak – 1 for all natural numbers k ≥ 2. Show that an = 3.7n–1 for all natural numbers.


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.


State whether the following statement is true or false. Justify.

Let P(n) be a statement and let P(k) ⇒ P(k + 1), for some natural number k, then P(n) is true for all n ∈ N.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×