1. tính:
a) 1.3+2.4+3.5+4.6+...+n.(n+2)
b) 1.5+2.6+3.7+...+n.(n+4)
c) 12 + 32+52+...+(2n+1)2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
ND
Nguyễn Đức Trí
VIP
25 tháng 7 2023
Bài 1 :
\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)
Ta thấy :
\(3=2^2-1\)
\(15=4^2-1\)
\(35=6^2-1\)
.....
\(9999=100^2-1\)
\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)
\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)
\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)
11 tháng 12 2023
a:
\(1^2+2^2+3^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\left(1\right)\)
Đặt \(S=1^2+2^2+...+n^2\)
Với n=1 thì \(S_1=1^2=1=\dfrac{1\left(1+1\right)\left(2\cdot1+1\right)}{6}\)
=>(1) đúng với n=1
Giả sử (1) đúng với n=k
=>\(S_k=1^2+2^2+3^2+...+k^2=\dfrac{k\left(k+1\right)\left(2k+1\right)}{6}\)
Ta sẽ cần chứng minh (1) đúng với n=k+1
Tức là \(S_{k+1}=\dfrac{\left(k+1+1\right)\cdot\left(k+1\right)\left(2\cdot\left(k+1\right)+1\right)}{6}\)
Khi n=k+1 thì \(S_{k+1}=1^2+2^2+...+k^2+\left(k+1\right)^2\)
\(=\dfrac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1\right)^2\)
\(=\left(k+1\right)\left(\dfrac{k\left(2k+1\right)}{6}+k+1\right)\)
\(=\left(k+1\right)\cdot\dfrac{2k^2+k+6k+6}{6}\)
\(=\left(k+1\right)\cdot\dfrac{2k^2+3k+4k+6}{6}\)
\(=\dfrac{\left(k+1\right)\cdot\left[k\left(2k+3\right)+2\left(2k+3\right)\right]}{6}\)
\(=\dfrac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)
\(=\dfrac{\left(k+1\right)\left(k+1+1\right)\left[2\left(k+1\right)+1\right]}{6}\)
=>(1) đúng
=>ĐPCM
b: \(A=1\cdot5+2\cdot6+3\cdot7+...+2023\cdot2027\)
\(=1\left(1+4\right)+2\left(2+4\right)+3\left(3+4\right)+...+2023\left(2023+4\right)\)
\(=\left(1^2+2^2+3^2+...+2023^2\right)+4\left(1+2+2+...+2023\right)\)
\(=\dfrac{2023\cdot\left(2023+1\right)\left(2\cdot2023+1\right)}{6}+4\cdot\dfrac{2023\left(2023+1\right)}{2}\)
\(=\dfrac{2023\cdot2024\cdot4047}{6}+\dfrac{2023\cdot2024}{1}\)
\(=2023\left(\dfrac{2024\cdot4047}{6}+2024\right)⋮2023\)
\(A=\dfrac{2023\cdot2024\cdot4047}{6}+2023\cdot2024\)
\(=2024\left(2023\cdot\dfrac{4047}{6}+2023\right)\)
\(=23\cdot11\cdot8\cdot\left(2023\cdot\dfrac{4047}{6}+2023\right)\)
=>A chia hết cho 23 và 11
DN
0
DT
Tính S = 1.3/3.5 + 2.4/5.7 + 3.5/7.9 + ... + ( n-1)( n+1) / (2n-1)(2n+1) + ... + 1002.1004/2005.2007
2
KN
13 tháng 8 2020
\(S=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)
\(\Rightarrow S=\frac{\left(2-1\right)\left(2+1\right)}{\left(2.2-1\right)\left(2.2+1\right)}+\frac{\left(3-1\right)\left(3+1\right)}{\left(3.2-1\right)\left(3.2+1\right)}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}\)
\(+..+\frac{\left(1003-1\right)\left(1003+1\right)}{\left(1003.2-1\right)\left(1003.2+1\right)}\)
\(\Rightarrow S=\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}\right)+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{3.2-1}-\frac{1}{3.2+1}\right)+...\)
\(+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+...+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{1003.2-1}-\frac{1}{1003.2+1}\right)\)
\(\Rightarrow S=1002.\frac{1}{4}-1002.\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}+\frac{1}{3.2-1}-...-\frac{1}{1003.2+1}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2007}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{2007}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}.\frac{668}{2007}\)
\(\Rightarrow S=\frac{501}{2}-\frac{27889}{223}\)
\(\Rightarrow S=125,4372197\)
\(\)
ND
Nguyễn Đức Trí
VIP
19 tháng 9 2023
a) Giả sử \(S_n=1^2+2^2+3^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\left(\forall n\inℕ^∗\right)\)
- Với \(n=1:\)
\(S_n=\dfrac{1.\left(1+1\right)\left(2.1+1\right)}{6}=\dfrac{2.3}{6}=1\left(luôn.đúng\right)\)
- Với \(n=k:\)
\(S_k=1^2+2^2+3^2+...+k^2=\dfrac{k\left(k+1\right)\left(2k+1\right)}{6}\left(\forall k\inℕ^∗\right)\left(luôn.đúng\right)\)
- Với \(n=k+1:\)
\(S_{k+1}=1^2+2^2+3^2+...+k^2+\left(k+1\right)^2\)
\(\Rightarrow S_{k+1}=\dfrac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1\right)^2\)
\(\Rightarrow S_{k+1}=\dfrac{k\left(k+1\right)\left(2k+1\right)+6\left(k+1\right)^2}{6}\)
\(\Rightarrow S_{k+1}=\dfrac{\left(k+1\right)\left[k\left(2k+1\right)+6\left(k+1\right)\right]}{6}\)
\(\Rightarrow S_{k+1}=\dfrac{\left(k+1\right)\left[2k^2+7k+6\right]}{6}\)
\(\Rightarrow S_{k+1}=\dfrac{\left(k+1\right)\left[2k^2+3k+4k+6\right]}{6}\)
\(\Rightarrow S_{k+1}=\dfrac{\left(k+1\right)\left[2k\left(k+\dfrac{3}{2}\right)+4\left(k+\dfrac{3}{2}\right)\right]}{6}\)
\(\Rightarrow S_{k+1}=\dfrac{\left(k+1\right)\left[\left(2k+4\right)\left(k+\dfrac{3}{2}\right)\right]}{6}\)
\(\Rightarrow S_{k+1}=\dfrac{\left(k+1\right)\left[\left(k+2\right)\left(2k+3\right)\right]}{6}\) (Đúng với \(n=k+1\))
Vậy \(S_n=1^2+2^2+3^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\left(\forall n\inℕ^∗\right)\left(dpcm\right)\)