[Đại 7] Toán dãy số

N

nhuquynhdat

$ B= 3+\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+3+...+100}$

$=3(1+\dfrac{1}{1+2}+ \dfrac{1}{1+2+3}+ \dfrac{1}{1+2+3+...+100}$

$=3(1+\dfrac{2}{3.2}+ \dfrac{2}{3.4}+...+ \dfrac{2}{100.101})$

$=3(1+\dfrac{1}{2}- \dfrac{1}{101})$

$=3. \dfrac{301}{202}$

$=\dfrac{903}{202}$
 
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T

tayhd20022001


1, Tính tổng $B=3+\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+3+...+100}$
$$Giải$$
Ta có :
$B=3+\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+3+...+100}$
\Rightarrow $B=3.1+\dfrac{1}{1+2}.3+\dfrac{1}{1+2+3}.3+...+ \dfrac{1}{1+2+3+...+100}.3$
\Rightarrow $B=3.(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+100})$
\Rightarrow $B=3.(1+\dfrac{2}{3.2}+\dfrac{2}{3.4}+...+\dfrac{2}{100.101})$
\Rightarrow $B=3.(1+\dfrac{1}{2}-\dfrac{2}{100.101})$
\Rightarrow $B=3.\dfrac{301}{202}$
\Rightarrow $B=\dfrac{903}{202}$
 
B

baochauhn1999

$ B= 3+\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+3+...+100}$

$=3(1+\dfrac{1}{1+2}+ \dfrac{1}{1+2+3}+ \dfrac{1}{1+2+3+...+100}$

$=3(1+\dfrac{2}{3.2}+ \dfrac{2}{3.4}+...+ \dfrac{2}{100.101})$

$=3(1+\dfrac{1}{2}- \dfrac{1}{101})$

$=3. \dfrac{301}{202}$

$=\dfrac{903}{202}$
 
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