1+[1+2]+[1+2+3]+...+[1+2+3+...+50]/1*50+2*49+...+50*1
Những câu hỏi liên quan
50-(1+1/3+........+1/50)=1/2+2/3+3/4+.......+49/50
chứng tỏ rằng : 49+48/2+47/3+...+2/48+1/49=50.(1/2+1/3+...+1/50)
tính các tổng sau
A=1*2+2*3+3*4+4*5+5*6+6*7...+49*50
B=1*50+2*49+3*48+...+49*2+50*1
Tính
1)A=1×2×3+2×3×4+.....+48×49×50
2)B=1×2+2×3+3×4+......+49×50
A = 1 × 2 × 3 + 2 × 3 × 4 + .....+ 48 × 49 × 50
ta có 4 x A = 1 x 2 x 3 x 4 + 2 x 3 x 4 x (5 -1) + .....+ 48 × 49 × 50 x (51 - 47)
= 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + ... + 48 x 49 x 50 x 51 - 47 x 48 x 49 x 50
= 48 x 49 x 50 x 51
suy ra A = (48 x 49 x 50 x 51) : 4
= 12 x 49 x 50 x 51
nhớ k cho mik nha rùi mik lm nốt cho
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Tính A=1*50+2*49+3*48+...+49*2+50*1
=> A = 1 x 50 + 2 x ( 50 - 1 ) + 3 x ( 50 - 2 ) + .... + 49 x ( 50 - 48 ) + 50 x ( 50 - 49 )
= 1 x 50 + 2 x 50 - 1 x 2 + 3 x 50 - 2 x 3 + ..... + 49 x 50 - 48 x 49 + 50 x 50 - 49 x 50
= ( 1 + 2 + 3 + .... + 50 ) x 50 + ( 1 x 2 + 2x 3 + .... + 49 x 50 )
= \(\frac{50.\left(50+1\right)}{2}\times50+\frac{49.50.51}{3}\)
= 63750 + 41650
= 105400
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-1 . 50 - 2 . 49 - 3 . 48 - ... - 49 . 2 - 50 . 1
Chứng tỏ:
1/26+1/27+...+1/49+1/50=99/50-97/49+...+7/4-5/3+3/2-1
Xét vế phải :
\(VT=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)
\(\text{Nhầm xíu , cho sửa lại nhé}\)
\(\text{Xét vế phải :}\)
\(VP=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)
\(A=\frac{\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}}{\frac{100}{1}+\frac{49}{2}+...+\frac{2}{49}+\frac{1}{50}}\)= ?
e, \(\frac{49}{1}+\frac{48}{2}+\frac{47}{3}+.......+\frac{2}{48}+\frac{1}{49}=50.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{50}\right)\)
\(\frac{49}{1}+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)
\(=1+1+...+1+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)(có 49 số 1)
\(=\left(1+\frac{48}{2}\right)+\left(1+\frac{47}{3}\right)+...+\left(1+\frac{2}{48}\right)+\left(1+\frac{1}{49}\right)+1\)
\(=\frac{50}{2}+\frac{50}{3}+...+\frac{50}{48}+\frac{50}{49}+\frac{50}{50}\)
\(=50\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)\)
Chúc bạn học tốt.
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