You are correct until the last line, which should be
1/2 [3k^2 + 7k + 6k + 10] = 1/2 (k+1) (3k+10)
Hello,
I start to solve this, but got stacked. Have no idea how to move on
Anybody help me ?
Code:5+8+11+…+ (3n+2) = 1/2n(3n+7) n=1 5+8+11+…+ (3*1+2) = 1/2 *1 (3*1+7) => 5=5 n=k 5+8+11+…+ (3k+2) = 1/2 k(3k+7) n=k+1 5+8+11+…+ (3(k+1)+2) = 1/2(k+1) (3(k+1)+7) 5+8+11+…+ (3k+5) = 1/2 (k+1) (3k+10) 5+8+11+…+ (3k+2) + (3k+5) = 1/2 (k+1) (3k+10) 1/2 k(3k+7) + (3k+5) = 1/2 (k+1) (3k+10) 1/2 [3k^2 + 10k + 5] = 1/2 (k+1) (3k+10) =>
1. I did not mention the base step here.
2. Guess that you did not understand me... I was working on the LHS untill I got k(k+3)(k+3)+4/4(k+1)(k+2)(k+3).... So the problem was that i didn't know how to transform k(k+3)(k+3)+4 to (k+1)(k+4), like on the RHS.
But after consultations with the professor, I've learn that
Thanks anyway
Hello everybody, again me
Now i'm trying to solve or prove this one:
1^2+2^2+3^2+....+n^2=n(n+1)(2n+1)/6
And I came to the third step where i got this one:
2k^3+10k^2+13k+6 on the LHS. I know that to solve this and get the same like on RHS, I need to work with division with polynoms, but even if i try to do that, still can't prove LHS=RHS.
Can anybody help me somehow?
Thanks