Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
a.
$S=3^0+3^2+3^4+...+3^{2002}$
$3^2S=3^2+3^4+3^6+...+3^{2004}$
$3^2S-S=(3^2+3^4+3^6+...+3^{2004})-(3^0+3^2+3^4+...+3^{2002})$
$8S=3^{2004}-3^0=3^{2004}-1$
$S=\frac{3^{2004}-1}{8}$
b.
$S=(3^0+3^2+3^4)+(3^6+3^8+3^{10})+....+(3^{1998}+3^{2000}+3^{2002})$
$=(3^0+3^2+3^4)+3^6(3^0+3^2+3^4)+....+3^{1998}(3^0+3^2+3^4)$
$=(3^0+3^2+3^4)(1+3^6+...+3^{1998})$
$=91(1+3^6+...+3^{1998})=7.13(1+3^6+...+3^{1998})\vdots 7$
Ta có đpcm.
b: \(S=\left(3^0+3^2+3^4\right)+...+3^{1998}\left(3^0+3^2+3^4\right)\)
\(=91\cdot\left(1+...+3^{1998}\right)⋮7\)
Ta có: \(S=1+3^2+3^4+3^6+...+3^{98}\)
\(=\left(1+3^2\right)+\left(3^4+3^6\right)+...+\left(3^{96}+3^{98}\right)\)
\(=10+3^4\cdot10+...+3^{96}\cdot10\)
\(=10\left(1+3^4+...+3^{96}\right)⋮10\)(ĐPCM)
\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 4
b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)
*Ta có: 32S=9S=32+34+.....+32002+32004
9S-3S=8S=32004-1=>S=\(\frac{3^{2004}-1}{8}\)
*S=(30+32+34)+(36+38+310)+.......+(31998+32000+32002)
=(30+32+34)+36(30+32+34)+..........+31998(30+3234)
=(30+32+34)(1+36+.....+31998)=91(1+36+...+31998) mà 91 chia hết cho 7
=>S chia hết cho 7
a,9S=32+34+36+................+32004
9S-S=(32+34+36+.............+32004)-(30+32+34+.............+32002)
8S=32004-30
S=\(\frac{3^{2004}-1}{8}\)
b,S=(30+32+34)+...........+(31998+32000+32002)
S=13.7+..................+31998.(1+32+34)
S=13.7+............+31998.13.7
S=(13+...........+31996.13).7 chia hết cho 7(đpcm)