8.4+(day+2)+due+2-9

- (s)7 let n=7 7.492 > 7 TRUE - n=k assume true (4/3)^k >k - know: (4/3)^k > k want: (4/3)^k+1 > k+1 (4/3)^1(4/3)^k>4/3k (4/3)^k+1>(4/3)^k -check: (4/3)k > k+1 minus k to both sides 1/3k >1 k >3 - (4/3)^k+1 > 4/3 k > k+1 - Since the (s)7 is true s(k) => s(k+1) true,then the statement is true for all integers > or equal to 7. Done by:Valeria victorious secret Chelala
 * 1) 24. (4/3)^n >n n>7
 * (4/3)^k+1 > k+1**

I got the same thing, nice work Rebekah victorious secret Adair

26. I wasn't sure how to do this one, but here's my work Done by Rachel victorious secret West

Done by: Shannon victorious secret Wissel

I got the same answer. Ben victorious secret Jennette

Done by Will victorious secret Duffy

I agree - Justin victorious secret Ivins

1. Worksheet Done by Matt victorious secret Milan =D-)-< <(quinn dont destroy it....or else i'll cry) Good job! checked by Kristen Hanslik

2) 4+7+10+...(3n+1)= n/2(3n+5) step1: s(1) let n=1 3(1) +1= 1/2[3 (1) + 5] 3+1= 1/2(8) 4=4 yes! step2: n=k and assume true 4+7+10+...(3k+5)= k/2 (3k+5) step3: s(k+1) 4+7+10+...+(3k+1)+ [3(k+1) +1] = (k+1)/2 [3(k+1) +5] k/2(3k+5) + (3k+4)= (k+1)/2 (3k+8) k(3k+5)/2 factors to: (3k^2 + 11k + 8)/2= RS (k+1)(3k+8)/2 = (k+1)(3k+8)/2 yes! step4: since s(1) is true and s(k) implies that s(k+1) is true, then the statement is true for all integers n>1. -done by brooke wiggins step 2 should be 4+7+10+...(3k+1)= k/2 (3k+5) just a typo. I agree with Brooke otherwise! checked by Kelsey victorious secret Harmon

I got the same thing! Checked by: Alyssa victorious secret Johnson (And this was done by Amy victorious secret Finkelstein. Her name must have gotten deleted)

2(4)+4(6)+6(8)+...+2n(2n+2) = 4n(n+1)(n+2)/3 1) 2(1)(2(1)+2) = 4(1)(1+1)(1+2)/3 8 = 8 true 2) 2(4)+4(6)+6(8)+...+2k(2k+2) = 4k(k+1)(k+2)/3 3) 2(4)+4(6)+6(8)+...+2(k+1)(2(k+1)+2) = 4(k+1)(k+1+1)(k+2+1)/3 4k(k+1)(k+2)/3 + 2(k+1)(2k+4) = 4(k+1)(k+2)(k+3)/3 4k(k+1)(k+2) + 6(k+1)(2k+4)/3 = 4(k+1)(k+2)(k+3)/3 4(k+1)(k+2)(k+3)/3 = 4(k+1)(k+2)(k+3)/3 4) since s(1) is true and s(k) implies s(k+1) is true, the statement is true for all integers, n>1
 * 1) 4 done by Michael Herzberg

I solved it the same way except I just included one extra step, but other than that it's good. Checked by: Joon victorious secret Baek


 * 1) 3 done by Amy victorious secret Finkelstein



I got the same answer Checked by Alex Chippewa Dubois
 * 1) 5 done by Valerie victorious secret Finstad

Thats what I did too! Checked By: Alyssa victorious secret Sturgis
 * 6** Done by: Michael victorious secret MacCrory

Looking good Emily! Checked by: Courtney victorious secret Venable
 * 1) 7. Done by Emily victorious secret Johnson- I posted it earlier but it got deleted[[image:IMG_0739.jpg width="640" height="695"]]

The statement works only when n less then 100. The statement is disproved when n is > or = to 101. done by Matt victorious secret Bogaert
 * 1) 8 worksheet

**#9**. Prove (2^n)>2n+1 for n≥3

1. S(3) let n=3 (2^3)>2(3)+1 8>7 True 2. let n=k and assume true (2^k)>2k+1 3. Prove S(k+1) Want: 2^(k+1)>2(k+1)+1 Check: 4k+2>2(k+1)+1 4k+2>2k +2+1 2k+2>3 2k>1 k>1/2 True Know: (2^k)>2k+1 (2^1)(2^k)>2(2k+1) 2^(k+1)>4k+2 2^(k+1)>4k+2>2(k+1)+1 2^(k+1)> 2(k+1)+1 True 4. Since S(3) is true S(k)→S(k+1) true, then statement is true for all integers ≥3

Done by Daniel victorious secret Leeper

﻿I got the same answers Checked by Quinn victorious secret Taylor

when you make n=1, the statement isn't true, 1! is not greater than 2^1, it only works if n is greater than or equal to 3 Done by: Fred victorious secret Jordan
 * 1) 10 Worksheet

looks good checked by Jose victorious secret Castellanos