English

1 + 3 + 32 + ... + 3n−1 = 3 N − 1 2 - Mathematics

Advertisements
Advertisements

Question

1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]

 
Advertisements

Solution

Let P(n) be the given statement.
Now,

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

\[\text{ Step 1} : \]

\[P(1) = 1 = \frac{3^1 - 1}{2} = \frac{2}{2} = 1\]

\[\text{ Hence, P(1) is true } . \]

\[\text{ Step 2:}  \]

\[\text{ Let P(m) is true .}  \]

\[\text{ Then } , \]

\[1 + 3 + 3^2 + . . . + 3^{m - 1} = \frac{3^m - 1}{2}\]

\[\text{ We shall prove that P(m + 1) is true }  . \]

\[\text{ That is, }  \]

\[1 + 3 + 3^2 + . . . + 3^m = \frac{3^{m + 1} - 1}{2}\]

\[\text{ Now, we have: }  \]

\[1 + 3 + 3^2 + . . . + 3^{m - 1} = \frac{3^m - 1}{2}\]

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

\[ \Rightarrow 1 + 3 + 3^2 + . . . + 3^m = \frac{3^m - 1 + 2 \times 3^m}{2} = \frac{3^m (1 + 2) - 1}{2} = \frac{3^{m + 1} - 1}{2}\]

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

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 12: Mathematical Induction - Exercise 12.2 [Page 27]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 3 | Page 27

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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


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

1.2 + 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

1.3 + 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

1/1.2.3 + 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

`1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/((2n + 1)(2n +3)) = n/(3(2n +3))`

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 ∈ N: 102n – 1 + 1 is divisible by 11


If P (n) is the statement "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.


\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]

 


\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]


\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\] 


1.2 + 2.22 + 3.23 + ... + n.2= (n − 1) 2n+1+2

 

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

 

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


a + ar + ar2 + ... + arn−1 =  \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]

 

52n −1 is divisible by 24 for all n ∈ N.


32n+2 −8n − 9 is divisible by 8 for all n ∈ N.


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

 

\[\frac{n^7}{7} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{n^2}{2} - \frac{37}{210}n\] is a positive integer for all n ∈ N.  

 


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

 


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


Prove that 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:

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)


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

Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1


Answer the following:

Prove by method of induction loga xn = n logax, x > 0, n ∈ N


Define the sequence a1, a2, a3 ... as follows:
a1 = 2, an = 5 an–1, for all natural numbers n ≥ 2.

Use the Principle of Mathematical Induction to show that the terms of the sequence satisfy the formula an = 2.5n–1 for all natural numbers.


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


Prove the statement by using the Principle of Mathematical Induction:

n3 – n is divisible by 6, for each natural number n ≥ 2.


Prove the statement by using the Principle of Mathematical Induction:

2n < (n + 2)! for all natural number n.


Prove the statement by using the Principle of Mathematical Induction:

`sqrt(n) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(n)`, for all natural numbers n ≥ 2.


A sequence b0, b1, b2 ... is defined by letting b0 = 5 and bk = 4 + bk – 1 for all natural numbers k. Show that bn = 5 + 4n for all natural number n using mathematical induction.


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)`.


If xn – 1 is divisible by x – k, then the least positive integral value of k is ______.


Consider the statement: “P(n) : n2 – n + 41 is prime." Then which one of the following is true?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×