✨ Best Answer ✨
階差数列は
3^2,5^2,7^2…(2n+1)^2→4n^2+4n+1
よって、
1+4/6n(2n^2-3n+1)+4/2n(n-1)+n-1
=2/3n(2n^2-3n+1)+2n(n-1)+n
=1/3n(4n^2-6n+2+6n-6+3)
=1/3n(4n^2-1)
続きが分からないです。(2)です。
✨ Best Answer ✨
階差数列は
3^2,5^2,7^2…(2n+1)^2→4n^2+4n+1
よって、
1+4/6n(2n^2-3n+1)+4/2n(n-1)+n-1
=2/3n(2n^2-3n+1)+2n(n-1)+n
=1/3n(4n^2-6n+2+6n-6+3)
=1/3n(4n^2-1)
Users viewing this question
are also looking at these questions 😉
ありがとうございます♪