Question

$\maltese$ Prove that the following equation is true for $\forall \: n \in {\mathbb{N}}$ by using Principle Of Mathematical Induction.

${(1)3 + (2)3^{2} + (3)3^{3} + \dots + (n)3^{n} = \dfrac{(2n - 1)3^{n + 1} + 3}{4} }$

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Answer 1

Step-by-step explanation:

Given :-

(1)3+(2)3²+(3)3³+...+(n)3^n =[(2n-1)(3)^(n+1) +3]/4

To find:-

Prove that by using the method Mathematical Induction ?

Solution :-

Given that :-

(1)3+(2)3²+(3)3³+...+(n)3^n =[(2n-1)(3)^(n+1) +3]/4

Put n = 1 then

(1)3¹ = [{(2×1)-1}(3)^(1+1) +3]/4

=>3 =[(2-1)(3)²+3]/4

=> 3 = ((1)9+3)/4

=> 3 = 12/4

=> 3 = 3

This statement is true for n = 1

Now ,

Let it is true for n = k then

Put n = k then

(1)3+(2)3²+(3)3³+...+(k)3^k =[(2k-1)(3)^(k+1) +3]/4

Now we need to prove that the result is true for

n = k+1

Put n = k+1 then

(1)3+(2)3²+(3)3³+...+(k+1)3^(k+1)

=[(2(k+1)-1)(3)^(k+1+1) +3]/4

RHS = [(2k+2-1)(3)^(k+2)+3]/4

=>RHS =[ (2k+1)(3)^(k+2) +3 ]/4

But by our supposition RHS = [(2k-1)(3)^(k+1) +3]/4

On adding (k+1)3^(k+1) both sides then

LHS = (1)3+(2)3²+(3)3³+...+(k)3^k+(k+1)(3)^(k+1)

RHS = [(2k-1)(3)^(k+1) +3]/4 + (k+1)(3)^(k+1)

=> [(2k-1)(3)^(k+1)+3+4(k+1)(3)^(k+1)]/4

=> (3)^(k+1)[(2k-1)+3+4(k+1)]/4

=> (3)^(k+1)[2k-1+3+4k+4]/4

=>(3)^(k+1)(6k+3+3)/4

=> (3)^(k+1)[3(2k+1)+3]/4

=> [(2k+1)(3)^(k+1+1)+3]/4

=> [(2k+1)(3)^(k+2)+3]/4

LHS = RHS is true for n = k+1

The result is true for n = k+1

This result is true for all positive integers for n .

Hence, Proved.

Used formulae:-

• a^m × a^n = a^(m+n)
• (a^m)^n = a^(mn)

Used Method:-

• Mathematical Induction.

Answer 2

Step-by-step explanation:

Given :

(1)3+(2)3²+(3)3³+...+(n)3^n =[(2n-1)(3)^(n+1) +3]/4

To find:

Prove that by using the method Mathematical Induction ?

Solution :

Given that :

(1)3+(2)3²+(3)3³+...+(n)3^n =[(2n-1)(3)^(n+1) +3]/4

Put n = 1 then

(1)3¹ = [{(2×1)-1}(3)^(1+1)

=>3 =[(2-1)(3)²+3]/4

=> 3 = ((1)9+3)/4

=> 3 = 12/4

=> 3 = 3

This statement is true for n = 1

Now,

Let it is true for n = k then

Put n = k then

(1)3+(2)3²+(3)3³+...+(k)3^k =[(2k-1)(3)^(k+1) +3]/4

Now we need to prove that the result is true for

n =k+1

Put n=k+1 then

(1)3+ (2)3²+ (3)3³+...+(k+1)3^(k+1)

=[(2(k+1)-1)(3)^(k+1+1) +3]/4

RHS = [(2k+2-1)(3)^(k+2)+3]/4

=>RHS =[(2k+1)(3)^(k+2) +3 ]/4

But by our supposition RHS = [(2k-1) (3)^(k+1) +3]/4

On adding (k+1)3^(k+1) both sides then

LHS = (1)3+(2)3²+(3)3³+...+(k)3^k+(k+1) (3)^(k+1)

RHS = [(2k-1)(3)^(k+1) +3]/4 + (k+1)(3)^(k+1) => [(2k-1)(3)^(k+1)+3+4(k+1)(3)^(k+1)]/4

=> (3)^(k+1)[(2k-1)+3+4(k+1)]/4

=> [(2k+1)(3)^(k+1+1)+3]/4

=> [(2k+1)(3)^(k+2)+3]/4

LHS = RHS is true for n=k+1

The result is true for n=k+1

This result is true for all positive integers for n.

Hence, Proved.

Used formulae:

a^m x a^n = a^(m+n) (a^m)^n = a^(mn)

Used Method:

Mathematical Induction.

Thanks

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