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# Did I do this math induction proof correctly? (9^n -1=8y)?

(This equation states that 9 raised to the power of n minus by 1 equals a multiple of 8.)

So, this is what I tried:

(prove it works for n=1)

9^1 -1=8=8y (where y is a variable integer and happens to equal 1 this time)

(assume it works for all values of n)

(prove it works for n+1)

9^n -1=8y

9(9^n -1)=9(8y)

9^(n+1) -9=72y

OR should the final step be:

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

9(9^n) -1=8y

9(8y +1) -1=8y

72y+8=8y

8(9y+1)=8y

Thank you everybody in advance for the help.  For n = 1, you prove it is correct.

let n = m ( m is any integral value for which the equation is assumed to be correct)

then, 9^m – 1 = 8y

Now we have to show that

9^(m+1) -1 = 8y'

=> 9^m . 9 - 1 = 8y'

=> (8y +1) . 9 - 1 = 8y'

=> 9. ( 8y ) + 9 - 1 = 8 y'

=> 8. ( 9y ) + 8 = 8 y'

=> 8. ( 9y + 1 ) = 8 y'

This proves it by method of induction.

n.b. y' is a constant like y. we have here,

9^(n -1) = 8y

n = 0 ==> 9^(0-1) = 8y ==> y = 1/72

n = 1 ==> 9^(1-1) = 8y ==> y = 1/8

n = 2 ==> 9^(2-1) = 8y ==> y = 9/8

n = 3 ==> 9^(3-1) = 8y ==> y = 81/8

n = 4 ==> 9^(4-1) = 8y ==> y = 729/8

. . . . . . . .

,. . . . . . .

. . . . and so on . . . . So assume that 9^n-1 = 8y. Consider 9^(n+1)-1. 9^(n+1)-1 = 9*(9^n)-1=9*9^n-9^n+9^n-1= 9^n(9-1)+9^n-1. By hypothesis 9^n-1 = 8y, so you have: 9^n*8+8y=8(9^n+y). Hence, if z=9^n+y, then 9^(n+1)-1=8z, and you have shown the result by induction.