問題詳情:
觀察下列各式: +3×1+1, +3×2+1, +3×3+1,
…
(1)猜想①= .
②= ,其中n爲正整數.
(2)請*②式.
(3)計算: +++…+.
【回答】
解:(1)①由題意可得,
=20152+3×2015+1,
故*爲:20152+3×2015+1;
②=n2+3n+1,
故*爲:n2+3n+1;
(2)*:∵()2=n4+6n3+11n2+6n+1,
(n2+3n+1)2=n4+6n3+11n2+6n+1,
∴=n2+3n+1;
(3)+++…+
=+…+
=
=
=
=
=
=.
知識點:二次根式的乘除
題型:解答題