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a) Tổng S có: (98-2):2+1=49 số hạng
b) S=22+24+26+....+298
=> 22A=22(22+24+26+....+298)
=> 4A=24+26+28+....+2100
=> 4A-A=(24+26+28+....+2100)-(22+24+26+....+298)
=> 3A=2100-22
=> \(A=\frac{2^{100}-2^2}{3}\)
4
a)\(S=1+2+2^2+...+2^{10}\)
\(2S=2+2^2+2^3+...+2^{11}\)
\(2S-S=\left(2+2^2+2^3+...+2^{11}\right)-\left(1+2+2^2+...+2^{10}\right)\)
\(S=2^{11}-1\)
b)\(S=1+3+3^2+...+3^6\)
\(3S=3+3^2+3^3+...+3^7\)
\(3S-S=\left(3+3^2+3^3+...+3^7\right)-\left(1+3+3^2+...+3^6\right)\)
\(2S=3^7-1\)
\(S=\frac{3^7-1}{2}\)
a.\(S=1+2+2^2+...+2^{10}\)
\(2S=2+2^2+2^3+...+2^{11}\)
\(\Rightarrow2S-S=S=\left(2+2^2+2^3+...+2^{11}\right)-\left(1+2+2^2+...+2^{10}\right)\)
\(=2^{11}-1\)
b) \(S=1+3+3^2+...+3^6\)
\(3S=3+3^2+3^3+...+3^7\)
\(\Rightarrow3S-S=2S=\left(3+3^2+3^3+...+3^7\right)-\left(1+3+3^2+...+3^6\right)\)
\(2S=3^7-1\Rightarrow S=\frac{3^7-1}{2}\)
B=2+8+16+32+64+128=(2+8)+(16+24)+(128+32)=10+40+160=210 CHIA HẾT CHO 5
B=2+8+16+32+64+128=250mà 250 chia het cho 5 vậy B chia hết cho 5
a/ Ta tính trường hợp tổng quát có n số hạng. Ta có:
+/ S1 = 1 + 2 + 3 + ....+n = \(\frac{n\left(n+1\right)}{2}\)
+/ S2 = 1.2 + 2.3 + 3.4 +...+ n(n+1)
3S2 = 1.2.3 + 2.3.3 + 3.4.3 +..+ n(n+1).3
3S2= 1.2.3 + 2.3.(4-1) + 3.4.(5-2) +..+ n(n+1)(n+2 -(n-1))
3S2= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +.. - (n-1)n(n+1) + n(n+1)(n+2)
3S2= n(n+1)(n+2)
=> S2 = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Tính S = 1² + 2² + ...+ n²
Ta có: S2 - S1 = [1.2 + 2.3 + 3.4 +...+ n(n+1)]-(1 + 2 + 3 + ....+n)
=> S2 - S1=(1.2-1)+(2.3-2)+(3.4-3)+...+[n(n+1)-n]
=> S2 - S1=1+4+9+...+n2=12+22+32+...+n2=S
Như vậy: S=S2-S1=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\)
=> \(S=n\left(n+1\right).\left(\frac{n+2}{3}-\frac{1}{2}\right)\)
=> \(S=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
Thay n=98 => \(S=\frac{98.99.197}{6}=318549\)
b/ 2014.2016=2014(2015+1)=2014+2014.2015=2014+2015(2015-1)=2014+20152-2015=20152-1<20152
Vậy 2014.2016<20152