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https://olm.vn/cau-hoi/a-cho-a12211216211002-ctr-a12-b-cho-p122132142120232-ctr-p-khong-la-so-tu-nhien-c-cho-c132152172120211.8293222842881
Cô làm rồi em nhá
Câu a, xem lại đề bài
Câu b:
P = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ...+ \(\dfrac{1}{2023^2}\)
Vì \(\dfrac{1}{2^2}\) < \(\dfrac{1}{1.2}\) = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)
\(\dfrac{1}{3^2}\) < \(\dfrac{1}{2.3}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)
\(\dfrac{1}{4^2}\) < \(\dfrac{1}{3.4}\) = \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)
........................
\(\dfrac{1}{2023^2}\) < \(\dfrac{1}{2022.2023}\) = \(\dfrac{1}{2022}\) - \(\dfrac{1}{2023}\)
Cộng vế với vế ta có:
0< P < 1 - \(\dfrac{1}{2023}\) < 1
Vậy 0 < P < 1 nên P không phải là số tự nhiên vì không tồn tại số tự nhiên giữa hai số tự nhiên liên tiếp
Câu c:
C = \(\dfrac{1}{3^2}\) + \(\dfrac{1}{5^2}\) + \(\dfrac{1}{7^2}\) + ....+ \(\dfrac{1}{2021^2}\) + \(\dfrac{1}{2023^2}\) = C
B = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+.......+ \(\dfrac{1}{2020^2}\) + \(\dfrac{1}{2023^2}\) > 0
Cộng vế với vế ta có:
C+B = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{5^2}\)+ \(\dfrac{1}{6^2}\)+...+ \(\dfrac{1}{2023^2}\) > C + 0 = C > 0
Mặt khác ta có:
1 > \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{2023^2}\) (cm ở ý b)
Vậy 1 > C > 0 hay C không phải là số tự nhiên (đpcm)
\(A=1^2+2^2+3^2+....+10^2\\ A=1^{ }+\left(1+1\right)\cdot2+3\cdot\left(2+1\right)+.....+10\cdot\left(9+1\right)\\ A=1+2\cdot1+2+3\cdot2+3+....+10\cdot9+10\\ A=\left(1+2+3...+10\right)+\left(1\cdot2+3\cdot2+.....+10\cdot9\right)\)
Gọi 1+2+3+...+10 là P
Số số hạng là: (10 - 1) : 1 +1 = 10 (số)
P = (10+1) . 10 : 2 = 55
P = 55
Gọi \(1\cdot2+2\cdot3+....+9\cdot10\) là C
\(C=1\cdot2+2\cdot3+....+9\cdot10\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot3+....+9\cdot10\cdot3\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+....+9\cdot10\cdot\left(11-8\right)\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+.....+9\cdot10\cdot11-8\cdot9\cdot10\\ 3\cdot C=9\cdot10\cdot11\\ 3\cdot C=990\\ C=330\)
\(=>A=P+C\\ =>A=55+330\\ A=385\)
b)
\(B=5^2+10^2+15^2+...+50^2\\ B=5^2+\left(2\cdot5\right)^2+\left(3\cdot5\right)^2+....+\left(5\cdot10\right)^2\\ B=5^2+2^2\cdot5^2+3^2\cdot5^2+...+5^2\cdot10^2\\ B=5^2\cdot\left(1+2^2+3^2+....+10^2\right)\\ B=25\cdot\left(1+2^2+3^2+....+10^2\right)\)
\(\left(1+2^2+3^2+....+10^2\right)=A\)
\(=>B=25\cdot A\\ B=25\cdot385\\ B=9625\)
Lời giải:
\(B=(1.2)^2+(2.2)^2+(3.2)^2+...+(10.2)^2\)
\(=2^2.1^2+2^2.2^2+2^2.3^2+...+2^2.10^2=2^2(1^2+2^2+...+10^2)\)
\(=4A=4.385=1540\)
Đặt A=11⋅2+12⋅3+...+17⋅8A=11⋅2+12⋅3+...+17⋅8
Dễ thấy: B=122+132+...+182B=122+132+...+182<A=11⋅2+12⋅3+...+17⋅8(1)<A=11⋅2+12⋅3+...+17⋅8(1)
Ta có:A=11⋅2+12⋅3+...+17⋅8A=11⋅2+12⋅3+...+17⋅8
=1−12+12−13+...+17−18=1−12+12−13+...+17−18
=1−18<1(2)=1−18<1(2)
Từ (1);(2)(1);(2) ta có: B<A<1⇒B<1
\(\dfrac{2^2}{1\times3}\times\dfrac{3^2}{2.4}\times\dfrac{4^2}{3.5}\times\dfrac{5^2}{4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.3.2.4.3.5.4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.2.3.3.4.4.5.2.3}=\dfrac{2^2.3^2.4^2.5^2}{3^3.2^2.4^2.5.1}=\dfrac{5}{3.1}=\dfrac{5}{3}\)
\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4.6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot2\cdot4^2\cdot4^2\cdot5\cdot6}\\ =\dfrac{2\cdot5}{6}=\dfrac{5}{3}\)
a)\(...A=\dfrac{2^{50+1}-1}{2-1}=2^{51}-1\)
b) \(...\Rightarrow B=\dfrac{3^{80+1}-1}{3-1}=\dfrac{3^{81}-1}{2}\)
c) \(...\Rightarrow C+1=1+4+4^2+4^3+...+4^{49}\)
\(\Rightarrow C+1=\dfrac{4^{49+1}-1}{4-1}=\dfrac{4^{50}-1}{3}\)
\(\Rightarrow C=\dfrac{4^{50}-1}{3}-1=\dfrac{4^{50}-4}{3}=\dfrac{4\left(4^{49}-1\right)}{3}\)
Tương tự câu d,e,f bạn tự làm nhé
12 + 52 + 62 = 1 + 25 + 36 = 62
22 + 32 + 72= 4 + 9 + 49 = 62
Vậy 12 + 52 + 62 = 22 + 32 + 72