Tính : B = 1^2 + 2^2 + ... + 99^2
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Ta có: B= \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\)
=> \(\frac{1}{2}B=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{100}+\left(\frac{1}{2}\right)^{100}\)
=> B - \(\frac{1}{2}B=\left(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\right)\)
\(-\left(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{4}\right)^4+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{100}+\left(\frac{1}{2}\right)^{100}\right)\)
=> B - \(\frac{1}{2}B=\left(\frac{1}{2}+\left(\frac{1}{2}\right)^{99}\right)-\left(\left(\frac{1}{2}\right)^{100}+\left(\frac{1}{2}\right)^{100}\right)=\frac{1}{2}\)
=> B \(\times\left(1-\frac{1}{2}\right)=\frac{1}{2}\)
=> B = 1
Câu này chắc chắn đúng luôn
B = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^99
=> 2B = 2(1/2 + 1/2^2 + 1/2^3 + ... + 1/2^99)
=> 2B = 1 + 1/2 + 1/2^2 +... + 1/2^98
=> 2B - B = 1+1/2+1/2^2+...+1/2^98 - (1/2+1/^2+1/2^3+...+1/2^99)
=> B = 1-1/2^99
Vậy B = 1 - 1/2^99
\(B=1^2+2^2+...+99^2\)
\(\Rightarrow B=1.\left(2-1\right)+2.\left(3-1\right)+...+99.\left(100-1\right)\)
\(\Rightarrow B=1.2-1+2.3-2+3.3-3+...+99.100-99\)
\(\Rightarrow B=\left(1.2+2.3+...+99.100\right)-\left(1+2+3+...+99\right)\)
Đặt A = 1.2 + 2.3 +...+99.100
C = 1 + 2 + 3 +...+99
Ta có :
A= 1.2 + 2.3 +... + 99.100
3A = 1.2.3 + 2.3.3 + .... +99.100.3
3A = 1.2.3 + 2.3.(4-1) + ... +99.100.(101-98)
3A = (1.2.3 + 2.3.4 + ... + 99.100.101) - (1.2.3 +2.3.4 + ... + 98.99.100)
3A = 99 . 100 . 101
=> A = 99.100.101/3 = 333300
C = 1+2+3+...+99
SSH của C là 99 - 1 + 1 = 99 (SH)
tổng C = \(\frac{\left(1+99\right).99}{2}=4950\)
=> B = 333300 - 4950 = 328350
328250