2015乘2014+2014乘2013+2013乘2012+......+2乘1
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这是一类题:
1*2+2*3+•••n(n+1)=?
解法一:(裂项相消)
因
n(n+1) = 1/3[n(n+1)(n+2) -(n-1)n(n+1)]
所以
1*2+2*3+3*4......+N*(N+1) = 1/3(1*2*3 - 0) + 1/3(2*3*4 - 1*2*3)+1/3(3*4*5 - 2*3*4) +....+ 1/3[n(n+1)(n+2) -(n-1)n(n+1)]
= 1/3[(1*2*3 - 0)+(2*3*4 - 1*2*3)+(3*4*5 - 2*3*4)+....+[n(n+1)(n+2) -(n-1)n(n+1)]]
= 1/3n(n+1)(n+2)
解法二:(分别求和)
因 n(n+1) = n^2 + n
则
1*2+2*3+3*4......+N*(N+1)
= (1^2 + 2^2 +3^3+....+n^2) +(1+2+3+...+n)
= 1/6n(n+1)(2n+1) + 1/2n(n+1)
= 1/3n(n+1)(n+2)
把n=2014代入
结果为1/3*2014*2015*2016=2727117120
这是一类题:
1*2+2*3+•••n(n+1)=?
解法一:(裂项相消)
因
n(n+1) = 1/3[n(n+1)(n+2) -(n-1)n(n+1)]
所以
1*2+2*3+3*4......+N*(N+1) = 1/3(1*2*3 - 0) + 1/3(2*3*4 - 1*2*3)+1/3(3*4*5 - 2*3*4) +....+ 1/3[n(n+1)(n+2) -(n-1)n(n+1)]
= 1/3[(1*2*3 - 0)+(2*3*4 - 1*2*3)+(3*4*5 - 2*3*4)+....+[n(n+1)(n+2) -(n-1)n(n+1)]]
= 1/3n(n+1)(n+2)
解法二:(分别求和)
因 n(n+1) = n^2 + n
则
1*2+2*3+3*4......+N*(N+1)
= (1^2 + 2^2 +3^3+....+n^2) +(1+2+3+...+n)
= 1/6n(n+1)(2n+1) + 1/2n(n+1)
= 1/3n(n+1)(n+2)
把n=2014代入
结果为1/3*2014*2015*2016=2727117120
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