Question

I just need help checking my work please...this is a really big grade and I am not confident enough to submit my material...Please Help.

4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = (4(4n+1)(8n+7))/6
n=1 4*6 = (4(4(1)+1)(8(1)+7))/6
(4(5)(15))/6 =50 24≠50 the base situation doesn’t work, and thus it is proven false.

Answer 1

2. 12 + 42 + 72 + ... + (3n - 2)2 = (n(6n^2-3n-1))/2
n=1 12= ("1(" 〖6(1)〗^2-3(1)-1))/2 =(6-3-1)/2 =2/2 =1 thus 12=1 so it works for the basic situation n=1
n=k 12+42+72+…+(3k-2)2=(k(k^2-3k-1))/2 assume this is true.
n=k+1 12+42+72+…+(3k-2)2+(3k+1-2)2=((k+1)[6(k+1)^2-3(k+1)-1])/2

(k(6k^2-3k-1))/2 +(3k-2)2=((k+1)[6(k+1)^2-3(k+1)-1])/2
(6k^3-3k^2-k)/2 + ((6k-4)^2)/2 = ((k+1)[6(k+1)^2-3(k+1)-1])/2
6k3+15k2+11k+2=(k+1)(6k2+9k+2) I then used synthetic division to factor our k+1 and the results were equal. Thus the statement is proved.

Answer 2

3. For the given statement Pn, write the statements P1, Pk, and Pk+1.
2 + 4 + 6 + . . . + 2n = n(n+1)
P1=2=1(1+1)=2
Pk=2+4+6+…+2k=k(k+1)
Pk+1=2+4+6+…+2k+2(k+1)=(k+1)(k+1+1)