1*100+3*99+5*98+.+197*2+199*1=?

Question

1*100+3*99+5*98+.+197*2+199*1=?

Answer 1

1*100+3*99+5*98+.+197*2+199*1
=100+99+98+.+3+2+1
+99+98+.+3+2+1
+99+98+.+3+2+1
+98+.+3+2+1
+98+.+3+2+1
.
2+1
2+1
+1
+1
=[(100+99+98+.+2+1)+(99+98+.+2+1)+(2+1)+1]
+[(99+98+.+2+1)+(2+1)+1]
先计算(100+99+98+.+2+1)+(99+98+.+2+1)+(2+1)+1
(100+99+98+.+2+1)+(99+98+.+2+1)+(2+1)+1
=100x101/2+99x100/2+98x99/2+.+2x3/2+1x2/2
=(100x101+99x100+.+2x3+1x2)/2
=(100²+100+99²+99+98²+98+.+2²+2+1²+1)/2
=(1+2+...+99+100+1²+2²+.+99²+100²)/2（利用平方和公式）
=100x101/2+100x(100+1)(2x100+1)/6
=5050+338350
=343400
再计算(99+98+.+2+1)+(2+1)+1
(99+98+.+2+1)+(2+1)+1
=99x100/2+98x99/2+.+2x3/2+1x2/2
=(99x100+.+2x3+1x2)/2
=(99²+99+98²+98+.+2²+2+1²+1)/2（同上面的过程）
=(1+2+...+99+1²+2²+.+99²)/2
=99x100/2+99x(99+1)(2x99+1)/6
=4950+328350
=333300
∴1*100+3*99+5*98+.+197*2+199*1
=[(100+99+98+.+2+1)+(99+98+.+2+1)+(2+1)+1]
+[(99+98+.+2+1)+(2+1)+1]
=343400+333300
=676700
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