Please help me with this use the proving mathematical Induction

Question

7•8+7•8²+7•8³+....+7•8ⁿ=8(8ⁿ-1)

Answer 1

7*8+7*8^2+7*8^3+...+7*8^n=8(8^[n]-1)

show its true for n=1

7*8=8*(8-1)=8*7

assume its true for n and show that it is true for n+1

it is true for n

show its true for n+1

show that:

7*8+7*8^2+7*8^3+...+7*8^n+7*8^(n+1)=8(8^[n+1]-1)

7(8+8^2+8^3+...+8^n+8^(n+1))=

7(8+8^2+8^3+...+8^n)+7(8^(n+1))=

we assumed that it was true for n, so...

8(8^n-1)+7(8^(n+1))=

8(8^n)-8+7(8^(n+1))=

8^(n+1)-8+7(8^(n+1))=

combine like terms

8(8^(n+1))-8=

factor out an 8

8(8^(n+1)-1)

Answer 2

I suppose what you seek is a derivation of the given equation.  (You did not ask a specific question.)  Start with the geometric series.

 

1 + x + x + ... + x^n-1 = (1 - xⁿ)/(1 - x)  [for x ≠ 1]

 

Multiply both sides by x.

 

x + x² + x³ + ... + xⁿ = x(1 - xⁿ)/(1 - x)

 

x + x² + x³ + ... + xⁿ = x(xⁿ - 1)/(x - 1)

 

Multiply both sides by 7, and substitute x = 8.

 

7(8 + 8² + 8³ + ... + 8ⁿ) = 7[8(8ⁿ - 1)/(8 - 1) = 7[8(8ⁿ - 1)/7]

 

7(8 + 8² + 8³ + ... + 8ⁿ) = 8(8ⁿ - 1)

Answer 3

7•8 + 7•8² + 7•8³ +....+ 7•8ⁿ

7•8(1 + 8¹ + 8² + ...+ 8^n-1)

 

The sequence inside the parentheses is a geometric sequence with n terms.  a₁ = 1 and a_n = 8^n-1.  The common ratio is r = 8.  The formula for the sum of a geometric sequence is:

 

S = a₁•( 1-rⁿ)/(1-r) = 1•(1-8ⁿ)/(1-8) = (1-8ⁿ)(-7) = (8ⁿ-1)/7

 

So we have:

 

7•8(1 + 8¹ + 8² + ...+ 8^n-1) = 7•8•(8ⁿ-1)/7 = 8(8ⁿ-1)