Muốn chứng minh 3/4+8/9+15/16+...+2499/2500 không phải số tự nhiên thì chứng minh nó nhỏ hơn 1

Ta có: \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}=\frac{1.3}{2^2}.\frac{2.4}{3^2}....\frac{49.51}{50^2}\)

\(=\frac{1.2....49}{2.3...50}.\frac{3.4...51}{2.3...50}=\frac{1}{50}.\frac{51}{2}=\frac{51}{100}

S=43​+98​+...+25002499​

\(= \frac{2^{2} - 1}{2^{2}} + \frac{3^{2} - 1}{3^{2}} + . . . + \frac{5 0^{2} - 1}{5 0^{2}}\)

\(= \left(\right. 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right)\)

\(= 49 - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right)\)

\(\frac{1}{2^{2}} < \frac{1}{1 \cdot 2} = 1 - \frac{1}{2}\)

\(\frac{1}{3^{2}} < \frac{1}{2 \cdot 3} = \frac{1}{2} - \frac{1}{3}\)

...

\(\frac{1}{5 0^{2}} < \frac{1}{49 \cdot 50} = \frac{1}{49} - \frac{1}{50}\)

Do đó: \(\frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + . . . + \frac{1}{49} - \frac{1}{50} = 1 - \frac{1}{50}\)

=>\(\frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} < 1\)

=>\(0 < \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} < 1\)

=>\(0 > - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right) > - 1\)

=>\(0 + 49 > - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right) + 49 > - 1 + 49\)

=>49>B>48

=>S không là số tự nhiên

S=43​+98​+1615​+...+50002499​

\(S = 1 - \frac{1}{4} + 1 - \frac{1}{9} + 1 - \frac{1}{16} + . . . + 1 - \frac{1}{5000}\)

\(S = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{4} + + \frac{1}{9} + \frac{1}{16} + . . . + \frac{1}{5000} \left.\right)\)

\(S = 49 - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right) < 49\)\(\left(\right. 1 \left.\right)\)

Lại có : 

\(\frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} < \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{49.50}\)

\(= \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + . . . + \frac{1}{49} - \frac{1}{50} = 1 - \frac{1}{50} < 1\)

\(\Rightarrow\)\(- \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right) > - 1\)

\(\Rightarrow\)\(S = 49 - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right) > 49 - 1 = 48\)\(\left(\right. 2 \left.\right)\)

Từ (1) và (2) suy ra : 

\(48 < S < 49\)

Vậy S không là số tự nhiên 

Chúc các bạn học tốt nhé ! =))

\(\dfrac{n^2-1}{n^2}=1-\dfrac{1}{n^2}>1-\dfrac{1}{\left(n-1\right)n}\)

Từ đó ta có:

\(A=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+...+\dfrac{50^2-1}{50^2}>1-\dfrac{1}{1.2}+1-\dfrac{1}{2.3}+...+1-\dfrac{1}{49.50}\)

\(\Rightarrow A>49-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)

\(\Rightarrow A>49-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(\Rightarrow A>49-\left(1-\dfrac{1}{50}\right)=48+\dfrac{1}{50}>48\)

\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\\ A=\left(1+1+1+...+1\right)-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\\ A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\)

Có \(\dfrac{1}{4}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\\ \dfrac{1}{9}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\\ \dfrac{1}{16}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\\ ...\\ \dfrac{1}{2500}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1-\dfrac{1}{50}< 1\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1\)

\(\Rightarrow A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)>49-1\\ \Rightarrow A>48\)

B = 3/4 + 8/9 + 15/16 + .... + 2499/2500

B = (1 - 1/4) + (1 - 1/9) + (1 - 1/16) + ... + (1 - 1/2500)

B = (1 - 1/2²) + (1 - 1/3²) + (1 - 1/4²) + ... + (1 - 1/50²)

B = (1 + 1 + 1 + ... + 1) - (1/2² + 1/3² + 1/4² + ...+ 1/50²)

                49 số 1

B = 49 - (1/2² + 1/3² + 1/4² + ... + 1/50²)

=> B < 49 (1)

B > 49 - (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/49×50)

B > 49 - (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50)

B > 49 - (1 - 1/50)

B > 49 - 1 + 1/50

B > 48 + 1/50 > 48 (2)

Từ (1) và (2) => 48 < B < 49

=> B không phải là số nguyên ( đpcm)

B = 3/4 + 8/9+ 15/16 + ... + 2499/2500

B = (1 - 1/4) + (1 - 1/9) + (1 - 1/16) + ... + (1 - 1/2500)

B = (1 - 1/2²) + (1 - 1/3²) + (1 - 1/4²) + ... + (1 - 1/50²)

B = (1 + 1 + 1 + ... + 1) - (1/2² + 1/3² + 1/4² + .... + 1/50²)

              49 số 1

=> B = 49 - (1/2² + 1/3² + 1/4² + ... + 1/50²)

=> B < 49 (1)

B > 49 - (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/49×50)

B > 49 - (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50)

B > 49 - (1 - 1/50)

B > 49 - 1 + 1/50

B > 48 + 1/50 > 48 (2)

Từ (1) và (2) => 48 < M < 49

=> M không phải số nguyên ( đpcm)

A lộn, B không phải số nguyên nha