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a)\(2S=2\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(2S=2+1+...+\frac{1}{2^{99}}\)
\(2S-S=\left(2+1+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(S=2-\frac{1}{2^{100}}\)
phần b tương tự
a. S=1+1/2+1/2^2+1/2^3+...+1/2^100
2S=2+1+1/2+1/2^2+...+1/2^99
2S-S=(2+1+1/2+1/2^2+...+1/2^99)-(1+1/2+1/2^2+1/2^3+...+1/2^100)
S=2-1/2^100
S=2^101-1/2^100
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B=\(3^1+3^2+3^3+...+3^{300}\)
=\(\left(3^1+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{299}+3^{300}\right)\)
=\(3\left(1+3\right)+3^3\left(1+3\right)+...+3^{299}\left(1+3\right)\)
=\(3.4+3^3.4+...+3^{299}.4\)
=\(\left(3+3^3+...+3^{299}\right).4\)
Vì 4\(⋮\)2 mà trong một tích có 1 ts chia hết cho 2 thì tích đó chia hết cho 2 \(\Rightarrow\)B\(⋮\)2
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b=31+32+...+3300
b=(3+32)+(33+34)+...+(3299+3300)
b=3(1+3)+33(1+3)+...+3299(1+3)
b=4(3+33+...+3299)
b=2.2(3+33+...+3299)
\(\Rightarrow\)b\(⋮\)2
Vậy...
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a,B=3+32+33+34+...+3300
=>3B=32+33+34+...+3301
=>3B-B=(32+33+34+...+3301)-(3+32+33+34+...+3300)
=>2B=3301-3
=>B=3101-3/2
b,ta có:2B+3=3101-3+3=3101=3n
=>n=101
vậy n=101
l-i-k-e cho mình nha
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B= 3+ 3\(^2\)+ 3\(^3\)+ 3\(^4\)+...+ 3\(^{300}\).
B có số số hạng là:
( 300- 1): 1+ 1= 300( số)
Ta ghép 2 số 1 thì nhóm được số nhóm là:
300: 2= 150( nhóm)
=> B=( 3+ 3\(^2\))+ 3\(^3\)( 3+ 3\(^2\))+...+ 3\(^{299}\)( 3+ 3\(^2\)).
B= 11+ 3\(^3\)x 11+...+ 3\(^{299}\)x 11.
B= 11( 1+ 3\(^3\)+...+ 3\(^{299}\)).
Vậy B\(⋮\) 11.