设f(n)=cosnπ/2,则f(25)+f(26)+f(27)+···+f(42)=

f(n) + f(n+1) + f(n+2)+f(n+3)
= [f(n) + f(n+2)] + [f(n+1) + f(n+3)]
= [cosnπ/2 + cos(n+2)π/2] + [cos(n+1)π/2 + cos(n+3)π/2]
= cosnπ/2 + cos(nπ/2 + π) + cos(n+1)π/2 + cos[(n+1)π/2 + π]
= cosnπ/2 - cosnπ/2 + cos(n+1)π/2 - cos(n+1)π/2
= 0

即任意连续4项的和为0

所以
f(25)+f(26)+f(27)+···+f(42)
= f(41) + f(42)
= cos41π/2 + cos42π/2
= cosπ/2 + cosπ
= 0 - 1
= -1

由f(n)=cosnπ/2可知：
f(4n)=cos2nπ=1
f(4n+1)=cos(2nπ+π/2)=0
f(4n+2)=cos(2nπ+π)=-1
f(4n+3)=cos(2nπ+π+π/2)=0
所以
f(25)+f(26)+f(27)+···+f(42)
=(0-1+0+1)+……(0-1+0+1)=0