Răspuns:
Explicație pas cu pas:
a) E₁ = (81⁸ + 9¹⁶):(9¹⁵+27¹⁰+7·3³⁰) =
=[(3⁴)⁸ + (3²)¹⁶] : [(3²)¹⁵+(3³)¹⁰+7·3³⁰)] =
= (3⁴ˣ⁸+3²ˣ¹⁶):(3²ˣ¹⁵+3³ˣ¹⁰+7·3³⁰) =
= (3³²+3³²) : (3³⁰+3³⁰+7·3³⁰ ) =
= (3³²·2):[3³⁰·(1+1+7)] =
= (2·3³²) : (9·3³⁰) = (2·3³²):(3²·3³⁰) = (2·3³²):2³² = 2
E₁ = 2
b) E₂ = (1·2+2·3+3·4+.....+9·10) - (1²+2²+3²+.....+9²) =
= (2+6+12+20+30+42+56+72+90)-(1+4+9+16+25+36+49+64+81) =
= 330 - 285 = 45
sau
1·2+2·3+3·4+.....+n·(n+1) = n·(n+1)·(n+2)/3
1·2+2·3+3·4+.....+9·10 = 9·10·11/3 = 3·110 = 330
1²+2²+3²+......+n² = n·(n+1)(2n+1)/6
1²+2²+3²+......+9² = 9·10·19/6 = 15·19 = 285
330-285 = 45
E₂ = 45
