求幂级数+∞∑n=0n3−5n+1(2n+1)(n+2)x2n+2
的和函数 S(x).
答案 11x4−13x2+44(x2−1)2+94xarctanx+ln(1−x2)x2.
解析 根据题意,有+∞∑n=0n3−5n+1(2n+1)(n+2)x2n+2=+∞∑n=0(−54+n2−1n+2+94(2n+1))x2n+2=−5x24(1−x2)+x42(x2−1)2−(−1−ln(1−x2)x2)+94xarctanhx=11x4−13x2+44(x2−1)2+94xarctanx+ln(1−x2)x2.