English

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

Advertisements
Advertisements

Question

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

 
Advertisements

Solution

Let P(n) be the given statement.
Now,

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

\[\text{ Step 1} : \]

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

\[\text{ Thus, P(1) is true }  . \]

\[\text{ Step 2 } : \]

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

\[\text{ Then, } \]

\[1 . 2 + 2 . 2^2 + . . . + m . 2^m = (m - 1) 2^{m + 1} + 2\]

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

\[\text{ That is, } \]

\[1 . 2 + 2 . 2^2 + . . . + (m + 1) 2^{m + 1} = m . 2^{m + 2} + 2\]

\[\text{ Now, } \]

\[P(m) = 1 . 2 + 2 . 2^2 + . . . + m . 2^m = (m - 1) 2^{m + 1} + 2\]

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

\[ \Rightarrow P(m + 1) = 2m . 2^{m + 1} + 2 = m . 2^{m + 2} + 2\]

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

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

 

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

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 10 | Page 27

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

1.3 + 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+3/1)(1+ 5/4)(1+7/9)...`(1 + ((2n + 1))/n^2) = (n + 1)^2`

 

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

`1/1.4 + 1/4.7 + 1/7.10 + ... + 1/((3n - 2)(3n + 1)) = n/((3n + 1))`


Prove the following by using the principle of mathematical induction for all n ∈ Nn (n + 1) (n + 5) is a multiple of 3.


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


Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n– 9 is divisible by 8.


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)?


Given an example of a statement P (n) such that it is true for all n ∈ N.

 

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

 

1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.

 

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

 

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

 


n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.

 

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

 

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

 


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


\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{  is true for all n }  \in N?\]


\[\text { A sequence  } x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_1 = 2 \text{ and }  x_k = \frac{x_{k - 1}}{k} \text{ for all natural numbers } k, k \geq 2 . \text{ Show that }  x_n = \frac{2}{n!} \text{ for all } n \in N .\]


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


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

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


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, for all n ∈ N

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


Answer the following:

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

`1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "n"/(("n" + 2)("n" + 3)("n" + 4)) = ("n"("n" + 1))/(6("n" + 3)("n" + 4))`


Answer the following:

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


Answer the following:

Prove by method of induction

`[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)], ∀  "n" ∈ "N"`


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

1 + 3 + 5 + ... + (2n – 1) = n2 


Prove by induction that for all natural number n sinα + sin(α + β) + sin(α + 2β)+ ... + sin(α + (n – 1)β) = `(sin (alpha + (n - 1)/2 beta)sin((nbeta)/2))/(sin(beta/2))`


A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P(5) is true. On the basis of this he could conclude that P(n) is true ______.


Prove the statement by using the Principle of Mathematical Induction:

For any natural number n, 7n – 2n is divisible by 5.


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:

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


Prove the statement by using the Principle of Mathematical Induction:

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


If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×