Răspuns:
Explicație pas cu pas:
1/(2n-1)(2n+1)=1/2(2n-1)-1/2(2n+1)
1/1·3=1/2-1/2·3 Daca adunam parte cu parte, se vor reduce
1/3·5=1/2·3-1/2·5 termenii pe diagonala
1/(2n-3)(2n-1)=1/2(2n-3-1/2(2n-1)
1/(2n-1)(2n+1)=1/2(2n-1)-1/2(2n+1)
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∑=1/2-1/2(2n+1)=(2n+1-1)/2(2n+1)=2n/2(2n+1)=n/(2n+1)
c) se scrie termenul general ca diferente de fractii si la adunare s evor reduce termenii
1/1·4+1/4·7+.......+1/(3n-2)(3n+1)=
1/(3k-2)·(3k+2)=1/3(3k-2)-1/4(3k+1)
1/1·4=1/3-1/3·4
1/4·7=1/3·4-1/3·7
1/(3n-2)(3n+1)=1/3(3n-2)-1/3((3n+1)
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∑ = 1/3-1/3(3n+1)=(3n+1-1)/3(3n+1)=n/(3n+1)
