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Recursionhard
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The Legendre polynomials Pn(x)P_n(x)Pn​(x) satisfy: (n+1)Pn+1(x)=(2n+1)xPn(x)−nPn−1(x)(n+1)P_{n+1}(x) = (2n+1)xP_n(x) - nP_{n-1}(x)(n+1)Pn+1​(x)=(2n+1)xPn​(x)−nPn−1​(x) with P0(x)=1,P1(x)=xP_0(x) = 1, P_1(x) = xP0​(x)=1,P1​(x)=x. Compute P2(1)P_2(1)P2​(1).