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已知数列ａｎ =ｐｘｎ+ ｑｙｎ,其中 ｐ、ｑ、ｘ、ｙ∈Ｒ ,ｎ∈Ｎ ,则有ａｎ+2 =(ｘ+ ｙ)ａｎ+1-ｘｙａｎ. 证明 ∵ａｎ =ｐｘｎ + ｑｙｎ,∴ (ｘ+ ｙ)ａｎ+1-ｘｙａｎ=(ｘ + ｙ) (ｐｘｎ+1+ ｑｙｎ+1) -ｘｙ(ｐｘｎ + ｑｙｎ)=ｐｘｎ+2 + ｑｘｙｎ+1+ ｐｘｎ+1ｙ + ｑｙ
Known series an = pxn + qyn, where p, q, x, y ∈ R, n ∈ N, then there is an + 2 = (x + y) an + 1 - xyan. Proof ∵ an = pxn + qyn, ∴ (x+ y)an+1-xyan=(x + y) (pxn+1+ qyn+1) -xy(pxn + qyn)=pxn+2 + qxyn+1+ pxn+1y + qy