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# proof by induction

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4 for A i subbed in the values and expanded the brackets which gave me

15^n+1 - 8^n-1 - (120^n -8^n-1)

205^n then i divided that by 7 which gave me 7*15^n   7=k

B is what im confused about

so try for n =2

15^2 - 8^2-2 = 224  divisible 7 =32

assume true for n =k

15^k - 8^k-2

try for n=k+1

15^k+1 - 8^k-1

then i did 15*15^k - 8^k/8 (im not sure whether i can do this)

225^k - 1^k =224^k

224^k / 7 = 15^k

= 15^k * 7

Jan 10, 2019
edited by YEEEEEET  Jan 10, 2019
edited by YEEEEEET  Jan 10, 2019

#1
+2

b)    Show that this is true for n = 2

15^2 - 8^(2 - 2)   =  225  - 1  =  224   = 32(7)

Assume it is true for n =  k.....that is

15^k - 8^(k - 2)    is a multiple of 7

Prove it is true for  k + 1

15^(k + 1)  - 8^(k + 1 - 2)

15^(k + 1) - 8^( k - 1)    note YEEEEEET  that we can write this as

(14 + 1)^(k + 1) - (7 + 1)^(k - 1)          use the binomial theorem and expand

[14^(k + 1) + C(k + 1, 1)*14^k  + ... + 14 + 14^0* 1^(k + 1) ] -

[ 7^(k - 1 )  + C(k - 1, 1) * 7^(k - 2) + ....+7 + 7^0*1^ (k - 1) ]

The  terms   14^0*1^(k +1)   and  7^0*1^(k - 1)    just equal 1  and will "cancel" with the subtraction of the second expansion from the first

So...we have this simplification....

[ 14^(k + 1) + C(k + 1, 1)* 14^k  +...+ 14  - 7^(k - 1) - C(k - 1, 1)*7^(k - 2 ) -...- 7 ]  =

[ (7 *2)^(k + 1) + C( k + 1,1)* (7*2)^k + ....+ 14 - 7^( k - 1) - C(k - 1, 1)*7^(k - 2) -....- 7 ]

Every term in the expansion simplification will be divisible by 7, hence the result is a multiple of 7   Jan 10, 2019
#2
+1

oh wow i didnt know you can do binomial expansion with algebraic power

thank you

YEEEEEET  Jan 10, 2019
edited by YEEEEEET  Jan 10, 2019
#2
0

oh wow i didnt know you can do binomial expansion with algebraic number

thank you

YEEEEEET  Jan 10, 2019
#4
+3

Here's an alternative approach: Jan 10, 2019
edited by Alan  Jan 10, 2019
edited by Alan  Jan 10, 2019