1−1−4x2x=∑n=0∞1n+1(2nn)xn11−4x=∑n=0∞(2nn)xn
11−4x(1−1−4x2x)k=∑n=0∞(2n+kn)xn(1−1−4x2x)k=∑n=0∞kn+k(2n+k−1n)xn
∑k=0n(2kk)(2(n−k)n−k)=4n∑k=0n(2kk)(2(n−k)+1n−k)=12(4n+1−(2(n+1)n+1))∑k=0n(2kk)(2(n−k)+2n−k)=4n+1−(2(n+2)n+2)∑k=0n(2k+1k)(2(n−k)+1n−k)=4n+1−(2(n+2)n+2)
∑k=0n((2k+1k)+(2k+2k))=12(2(n+2)n+2)−1∑k=0n2n−k(2k+2k)=12(2(n+2)n+2)−2n+1∑k=0n3n−k(2k+3k)=12(2(n+3)n+3)−3n+2∑k=0nan−k(2k+4k)=12((2(n+4)n+4)−an+4+2an+3)∑k=0nbn−k(2k+5k)=12((2(n+5)n+5)−bn+5+3bn+4−bn+3)an:=(2+2)n+1−(2−2)n+122bn:=(5+52)n+1−(5−52)n+15
∑k=0n3k+3(2k+2k)=1n+3(2(n+2)n+2)−1∑k=0n2n−k4k+4(2k+3k)=1n+4(2(n+3)n+3)−2n+2∑k=0nan−k5k+5(2k+4k)=1n+5(2(n+4)n+4)+an+5−5an+4+5an+32∑k=0n3n−k+1−126k+6(2k+5k)=1n+6(2(n+5)n+5)−3n+4+12an:=(3+52)n+1−(3−52)n+15
∑k=0nk2k1n(2n−1−kn−1)=(2nn)∑k=0nk(k+1)2k1n(2n−1−kn−1)=4n∑k=0n2k(2n−kn)=4n
[xnym]11−4x1−4y(1−4x+1−4y)=14(2(n+m+1)n+m+1)∑k=0⌊n2⌋1(k!)22n−2k(n−2k)!=(2nn)n!∑k=0n(2kk)k!⋅(−1)n−k(2(n−k)n−k)(n−k)!={0(n≡1(mod2))(nn/2)2n!(n≡0(mod2))
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