Question

$chapter$
Principle of Mathematical Induction.
$excercise \: 4.1$


$question$
prove the following by using the principal of mathematical induction of all n € N:

$1)$
$1 + 3 + 3 {}^{2} + - - 3n {} - 1 = \binom{3n - 1}{2}$






​

Answer 1

Appropriate Question :-

Prove by Principal of Mathematical Induction

$\rm \: 1 + 3 + {3}^{2} + - - - + {3}^{n - 1} = \dfrac{ {3}^{n} - 1}{2}$

$\large\underline{\sf{Solution-}}$

To prove that,

$\rm \: 1 + 3 + {3}^{2} + - - - + {3}^{n - 1} = \dfrac{ {3}^{n} - 1}{2}$

Let assume that

$\rm \: P(n) : \rm \: 1 + 3 + {3}^{2} + - - - + {3}^{n - 1} = \dfrac{ {3}^{n} - 1}{2}$

Step :- 1 For n = 1

$\rm \: P(1) : \rm \: 1 = \dfrac{ {3}^{1} - 1}{2}$

$\rm \: P(1) : \rm \: 1 = \dfrac{ 3 - 1}{2}$

$\rm \: P(1) : \rm \: 1 = \dfrac{2}{2}$

$\rm \: P(1) : \rm \: 1 = 1$

$\rm\implies \:\rm \: P(1) \: is \: true \: for \: n= 1$

Step : - 2 For n = k, Assume that P(n) is true, where k is some natural number.

$\rm \: P(k) : 1 + 3 + {3}^{2} + - - - + {3}^{k - 1} = \dfrac{ {3}^{k} - 1}{2}$

Step :- 3 For n = k + 1, we have to prove that P(n) is true.

$\rm \: P(k + 1) : 1 + 3 + {3}^{2} + - - - + {3}^{k - 1} + {3}^{k} = \dfrac{ {3}^{k + 1} - 1}{2}$

Now, Consider LHS

$\rm \: 1 + 3 + {3}^{2} + - - - + {3}^{k - 1} + {3}^{k}$

$\rm \: =  \:\dfrac{ {3}^{k} - 1}{2} + {3}^{k}$

$\rm \: =  \:\dfrac{ {3}^{k} - 1 + 2. {3}^{k} }{2}$

$\rm \: =  \:\dfrac{ {3}^{k}(1 + 2) - 1 }{2}$

$\rm \: =  \:\dfrac{ {3}^{k}.3 - 1 }{2}$

$\rm \: =  \:\dfrac{ {3}^{k + 1} - 1 }{2}$

Hence,

$\rm\implies \:\rm \: P(n) \: is \: true \: for \: n= k + 1$

Hence, By Principal of Mathematical Induction,

$\rm \: 1 + 3 + {3}^{2} + - - - + {3}^{n - 1} = \dfrac{ {3}^{n} - 1}{2}$

Answer 2

SOLUTION :

• please check the attached file