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Đặt P=31-1+32-1+33-1+34-1+...+3n-1
=>P=30+31+32+33+...+3n-1
=>3.P=31+32+33+34+...+3n
=>3.P-P=31+32+33+34+...+3n-30-31-32-33-...-3n-1
=>2.P=3n-30
=>2.P=3n-1
=>\(P=\frac{3^n-1}{2}\)
Lại có: S=1+2+5+14+...+\(\frac{3^{n-1}+1}{2}\)
=>\(S=\frac{3^{1-1}+1}{2}+\frac{3^{2-1}+1}{2}+\frac{3^{3-1}+1}{2}+\frac{3^{4-1}+1}{2}+...+\frac{3^{n-1}+1}{2}\)
=>\(S=\frac{3^{1-1}+1+3^{2-1}+1+3^{3-1}+1+3^{4-1}+1+...+3^{n-1}+1}{2}\)
=>\(S=\frac{\left(3^{1-1}+3^{2-1}+3^{3-1}+3^{4-1}+...+3^{n-1}\right)+\left(1+1+1+1+...+1\right)}{2}\)
=>\(S=\frac{P+1.n}{2}\)
=>\(S=\frac{\frac{3^n-1}{2}+n}{2}\)
=>\(S=\frac{\frac{3^n-1}{2}+\frac{2n}{2}}{2}\)
=>\(S=\frac{\frac{3^n-1+2n}{2}}{2}\)
=>\(S=\frac{3^n-1+2n}{4}\)
nhìn cái cuối là biết quy luật đó bạn :))
\(S=\frac{3^{1-1}+1}{2}+\frac{3^{2-1}+1}{2}+\frac{3^{3-1}+1}{2}+...+\frac{3^{n-1}+1}{2}\)
\(S=\frac{\left(3^0+3^1+....+3^{n-1}\right)+\left(1+1+1+...+1\right)}{2}\left(\text{ có n c/s 1}\right)\)
\(S=\frac{\frac{\left(3^n-1\right)}{2}+n}{2}=3^n-1+\frac{n}{2}\)
chỗ 30+31+...+3n-1 bn tự tính :))
S=(3^0+1/2)+(3^1/2+1/2)+(3^2/2+1/2)+....+(3^n-1/2+1/2)
=n*1/2+1/2*(3^0+3^1+3^2+...+3^n-1)
=n^2/2+(3^n-1/4)=3^n+2-1/4
~~~~~~~~~~~~~~~~~~~~~
Lời giải:
$n=1$ thì $S=0$ nguyên nhé bạn. Phải là $n>1$
\(S=1-\frac{1}{1^2}+1-\frac{1}{2^2}+1-\frac{1}{3^2}+...+1-\frac{1}{n^2}\)
\(=n-\underbrace{\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)}_{M}\)
Để cm $S$ không nguyên ta cần chứng minh $M$ không nguyên. Thật vậy
\(M> 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n(n+1)}=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n}-\frac{1}{n+1}\)
\(M>1+\frac{1}{2}-\frac{1}{n+1}>1\) với mọi $n>1$
Mặt khác:
\(M< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{(n-1)n}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}\)
\(M< 1+1-\frac{1}{n}< 2\)
Vậy $1< M< 2$ nên $M$ không nguyên. Kéo theo $S$ không nguyên.
\(S=1+2+5+14+...+\dfrac{3^{n-1}+1}{2};\left(n\in N\backslash\left\{0\right\}\right)\)
\(2S=2+4+10+28+....+\left(3^{n-1}+1\right)=S_1\)
\(2S=\left[1+1+....+n\right]+\left[1+3+9+..+3^{n-1}\right]\)
\(S_1=1+1+1+..+n=n\)
\(S_2=1+3+9+....+3^{n-1}\)
\(3S_2=3+9+...+3^n\)
\(3S_2-S_2=2S_2=3^n-1\Rightarrow S_2=\dfrac{3^n-1}{2}\)
\(S=\dfrac{s_1+s_2}{2}=\dfrac{n+\dfrac{3^n-1}{2}}{2}=\dfrac{3^n+2n-1}{4}\)