Answer to Question #163498 in Discrete Mathematics for Ultra

Question #163498

2-2.7+2.7^2-...+2(-7)^n = (1- (-7)^n+1)/4


1
Expert's answer
2021-02-24T07:03:35-0500

PROOF BY INDUCTION

let P(n) be 2 - 2*7 + 2*72 -...+ 2*(-7)n= "\\frac {1- (-7)^{(n+1)} } 4"

Basic step (n = 0):

2*(-7)0= 2*1= 2

"\\frac {1- (-7)^{(0+1)} } 4 = \\frac {1- (-7)^{1} } 4= \\frac {1- (-7) } 4 = \\frac {1+ 7 } 4= \\frac 8 4 = 2"

We then note P(0) is true, as both sides of the equations is equal to 2.

Induction step

let P(k) be true => 2 - 2*7 + 2*72 -...+ 2*(-7)k= "\\frac {1- (-7)^{(k+1)} } 4"

We need to proove that P(k+1) is also true.

2 - 2*7 + 2*72 -...+ 2*(-7)k + 2*(-7)k+1= "\\frac {1- (-7)^{(k+1)} } 4" + 2*(-7)k+1=

="\\frac {1- (-7)^{(k+1)} + 8* (-7)^{(k+1)} } 4" "\\frac {1- (-7)^{(k+1)} } 4 + \\frac {8* (-7)^{(k+1)} } 4 =\\frac {1- (-7)^{(k+1)} + 8* (-7)^{(k+1)} } 4" =

= "\\frac {1+(-1)* (-7)^{(k+1)} + 8* (-7)^{(k+1)} } 4 = \\frac {1+(-1+8)* (-7)^{(k+1)} } 4" =

="\\frac {1 + 7* (-7)^{(k+1)} } 4= \\frac {1 -(- 7)* (-7)^{(k+1)} } 4= \\frac {1 - (-7)^{(k+2)} } 4" =

= "\\frac {1 - (-7)^{(k+1)+1} } 4"

We then note that P(k+1) is also true.

Conclusion

By the principle of mathematical induction, P(n) is true for all nonnegative integers n.



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