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Ta có công thức : với n thuộc N* thì ta luôn có :
\(1+\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)+1}{n\left(n+2\right)}=\frac{n^2+2n+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
Áp dụng vào bài toán ta được :
\(P=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{49.51}\right)+\frac{2}{51}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.......\frac{50^2}{49.51}+\frac{2}{51}\)
\(=\frac{\left(2.3.4...50\right)\left(2.3.4...50\right)}{\left(1.2.3...49\right)\left(3.4.5....51\right)}+\frac{2}{51}\)
\(=\frac{50.2}{51}+\frac{2}{51}=\frac{102}{51}=2\)
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\(A=\) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.50}\)
A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
A = \(\frac{1}{1}-\frac{1}{51}=\frac{50}{51}\)
1+1/1.3= 2^2/1.3
1+1/2.4=9/2.4=3^2/2.4
1+1/3.5=16/3.5=4^2/3.5
...................................
1+1/49.51=2500/49.51=50^2/49.50
2^2.3^2. ... . 50^2/ 1.3.2.4.3.5.4.6. .... 49.51 +2/51
2.50/51 +2/51 =2