Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.. . 1) 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4( 4n + 2) = 4(4n+1)(8n+7)/6. . 2) 1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = n(6n^2-3n-1)/2

Answer 1

)to get the summation u do :
sigma(r=1 to n)[4r( 4r + 2)] , ans would be a cubic polynomial but you have given ans as square polynomial so its false.
OR u can check using values as n=1,2,3... and check its wrong.

2)sigma(1 to n)[(3r-2)^2] ans would be a cubic polynomial and the given ans is also so we need to check.
=sigma(1 to n)(9r^2-12r+4)
=[9*n(n+1)(2n+1)/6-12*n(n+1)/2+4n] {using sigma(r^2)=n(n+1)(2n+1)/6 ; sigma(r)=n(n+1)/2 sigma(1)=n}
=n(6n^2-3n-1)/2 hope it helps