Question

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

Answer 1

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.