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2014-10-18T19:08:24+03:00
[1/(1·2) + 1/(2·3) +1/(3·4) + ......+1/(99·100)]÷[1/3^4 ·3³/5³ ·5^5/3^5] = [1/1-1/2+1/2-1/3+1/3-1/4+.....+1/99 -1/100]÷5²/3^6=(1/1- 1/100) × 3^6/5² = 99/100 ·3^6/5² = (11·3^8 )/(2²·5^4 )