Lời giải:
\(A=1+3+(3^2+3^3+3^4+3^5)+(3^6+3^7+3^8+3^9)+...+(3^{46}+3^{47}+3^{48}+3^{49})\)

\(=4+3^2(1+3+3^2+3^3)+3^6(1+3+3^2+3^3)+....+3^{46}(1+3+3^2+3^3)\)

\(=4+3^2.40+3^6.40+....+3^{46}.40\)

\(=10(4.3^2+4.3^6+..+4.3^{46})+4\)

Vậy $A$ có tận cùng là $4$

Đây là toán lớp 3 á!!!!
Mà bn có vt sai đề bài ko? Mk tính ko ra

để mik xem lại

\(S=1+3+3^2+3^3+...+3^{48}+3^{49}.\)

\(S=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{48}+3^{49}\right)\)

\(S=1\left(1+3\right)+3^2\left(1+3\right)+..+3^{48}\left(1+3\right)\)

\(S=4\left(1+3^2+....+3^{48}\right)\)

\(\Rightarrow S⋮4\)

b, Có : \(S=1+3+3^2+3^3+...+3^{48}+3^{49}\)

\(\Rightarrow3S=3+3^2+3^3+...+3^{48}+3^{49}+3^{50}\)

=> 3S - S = ( 1 + 3 + 3² + 3³  + ..... + 3⁴⁸ + 3⁴⁹  ) - ( 3 + 3³ + 3³ + .. + 3⁴⁹ + 3⁵⁰)

\(\Rightarrow2S=3^{50}-1\)

\(\Rightarrow S=\frac{3^{50}-1}{2}\)

\(\Rightarrow3^{50}-1=\left(...9\right)-1=\left(...8\right)\)( tận cùng là 8 )

\(\Rightarrow S=\frac{3^{50}-1}{2}=\frac{....8}{2}=\left(...4\right)\)

=> S có tận cùng là 4 

a) \(S=1+3+3^2+3^3+...+3^{48}+3^{49}\)

\(S=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{48}+3^{49}\right)\)

\(S=4+\left(3^2.1+3^2.3\right)+...+\left(3^{48}.1+3^{48}.3\right)\)

\(S=4+3^2.\left(1+3\right)+...+3^{48}.\left(1+3\right)\)

\(S=1.4+3^2.4+...+3^{48}.4\)

\(S=\left(1+3^2+....+3^{48}\right).4⋮4\)

giải giùm mình câu c luôn với

a) \(S=1+3+3^2+3^3+...+3^{49}\)

\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{48}+3^{49}\right)\)

\(=1\left(1+3\right)+3^2\left(1+3\right)+...+3^{48}\left(1+3\right)\)

\(=1.4+3^2.4+...+3^{48}.4\)

\(=\left(3+1\right)\left(1+3^2+...3^{48}\right)=4\left(1+3^2+...+3^{48}\right)⋮4^{\left(đpcm\right)}\)

b) Ta có: \(S=1+3+3^2+3^3+...+3^{49}\)

\(3S=3+3^2+3^3+...+3^{49}+3^{50}\)

\(3S-S=2S=3^{50}-1\Rightarrow S=\frac{3^{50}-1}{2}\)

Ta thấy: \(3^{50}=3^{4.12}.3^2=\left(3^4\right)^{12}.3^2=81^{12}.9=...9\) (tận cùng là 9)

Suy ra \(3^{50}-1=\left(...9\right)-1=...8\) (tận cùng là 8)

Suy ra \(\Rightarrow S=\frac{3^{50}-1}{2}=\frac{\left(...8\right)}{2}=...4\Rightarrow S\) tận cùng là 4

a) \(S=1+3+3^2+3^3+...+3^{49}\)

\(S=\left(1+3\right)+\left(3^2+3^3\right)+....+\left(3^{48}+3^{49}\right)\)

\(S=4+\left(3^2.1+3^2.3\right)+....+\left(3^{48}.1+3^{48}.3\right)\)

\(S=4+3^2.\left(1+3\right)+...+3^{48}.\left(1+3\right)\)

\(S=1.4+3^2.4+...+3^{48}.4\)

\(S=\left(1+3^2+...+3^{48}\right).4⋮4\)

a) S = 1 + 3 + 3² +...+ 3⁴⁸ + 3⁴⁹

=> 3S = 3 + 3² + 3³ +...+ 3⁴⁸ + 3⁴⁹ + 3⁵⁰

=> 3S - S = 3⁵⁰ - 1

=> S = \(\frac{3^{50}-1}{2}\)

       Vậy S = \(\frac{3^{50}-1}{2}\)

b) Câu này hơi khó!

S = \(\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+...+\left(3^{48}+3^{49}\right)\)

= \(4+3^2.\left(1+3\right)+3^4.\left(1+3\right)+...+3^{48}.\left(1+3\right)\)

= \(4+3^2.4+3^4.4+...+3^{48}.4\)

= \(4.\left(1+3^2+3^4+...+3^{48}\right)\text{ chia hết cho 4}\)

=> S chia hết cho 4 (đpcm).

b. Chưa rõ.

c. S = \(1+3+3^2+3^3+...+3^{49}\)

=> 3S = \(3.\left(1+3+3^2+3^3+...+3^{49}\right)\)

=> 3S = \(3+3^2+3^3+3^4+...+3^{50}\)

=> 3S - S = \(\left(3+3^2+3^3+3^4+...+3^{50}\right)-\left(1+3+3^2+3^3+...+3^{49}\right)\)

=> 2S = \(3^{50}-1\)

=> S = \(\frac{3^{50}-1}{2}\left(\text{đpcm}\right)\).

minh hiền bạn làm đúng rùi mong bạn sớm làm được phần b chúc học giỏ

Bài 1:

\(A=7+7^3+7^5+...+7^{1999}\)

\(\Rightarrow A=\left(7+7^3\right)+\left(7^5+7^7\right)+...+\left(7^{1997}+7^{1999}\right)\)

\(\Rightarrow A=\left(7+343\right)+7^4\left(7+7^3\right)+...+7^{1996}\left(7+7^3\right)\)

\(\Rightarrow A=350+7^4.350+...+7^{1996}.350\)

\(\Rightarrow A=\left(1+7^4+...+7^{1996}\right).350⋮35\)

\(\Rightarrow A⋮35\left(đpcm\right)\)

b2:

a) \(S=1+3+3^2+...+3^{49}\)

\(\Rightarrow S=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{48}+3^{49}\right)\)

\(\Rightarrow S=\left(1+3\right)+3^2\left(1+3\right)+...+3^{48}\left(1+3\right)\)

\(\Rightarrow S=4+3^2.4+...+3^{48}.4\)

\(\Rightarrow S=\left(1+3^2+...+3^{48}\right).4⋮4\)

\(\Rightarrow S⋮4\left(đpcm\right)\)

c) \(S=1+3+3^2+...+3^{49}\)

\(\Rightarrow3S=3+3^2+3^3+...+3^{50}\)

\(\Rightarrow3S-S=\left(3+3^2+3^3+...+3^{50}\right)-\left(1+3+3^2+...+3^{49}\right)\)

\(\Rightarrow2S=3^{50}-1\)

\(\Rightarrow S=\frac{3^{50}-1}{2}\left(đpcm\right)\)

\(A=7+7^3+7^5+...+7^{1999}\)

\(=\left(7+7^3\right)+\left(7^5+7^7\right)+...+\left(7^{1997}+7^{1999}\right)\)

\(=\left(7+7^3\right)+7^4\left(7+7^3\right)+...+7^{1996}\left(7+7^3\right)\)

\(=350+7^4.350+...+7^{1996}.350\)

\(=350.\left(1+7^4+...+7^{1996}\right)\)

Vì \(350⋮35\) nên \(350.\left(7+7^4+...+7^{1996}\right)⋮35\)

Vậy \(A⋮35\)(đpcm)

2.

a, \(S=1+3+3^2+...+3^{49}\)

\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{48}+2^{49}\right)\)

\(=\left(1+3\right)+3^2\left(1+3\right)+...+3^{48}\left(1+3\right)\)

\(=4+3^2.4+...+3^{48}.4\)

\(=4\left(1+3^2+...+3^{48}\right)\)

Vì \(4⋮4\) nên \(4\left(1+3^2+...+3^{48}\right)⋮4\)

Vậy \(S⋮4\) (đpcm)

c, \(S=1+3+3^2+...+3^{49}\)

\(3S=3+3^2+3^3+...+3^{50}\)

\(3S-S=\left(3+3^2+3^3+...+3^{50}\right)-\left(1+3+3^2+...+3^{49}\right)\)

\(2S=3^{50}-1\)

\(S=\left(3^{50}-1\right):2\) (đpcm)

b, Ta có: \(S=\left(3^{50}-1\right):2\)

\(=\left(3^{48}.3^2-1\right):2\)

\(=\left[\left(3^4\right)^{14}.9-1\right]:2\)

\(=\left[\overline{\left(...1\right)}^{14}.9-1\right]:2\)

\(=\left[\overline{\left(...1\right)}.9-1\right]:2\)

\(=\left[\overline{\left(...9\right)}-1\right]:2\)

\(=\overline{\left(...8\right)}:2\)

\(=\overline{...4}\)

Vậy chữ số tận cùng của S là 4