Question

cho s bằng 1+3^1+3^2+3^3+....3^30

cho n bằng 1+4+4^2+....+4^132

Tính s,n

Answer 1

S = 1 + 3 + 3² + 3³ + ... + 3³⁰

3S = 3 + 3² + 3³ + 3⁴ + ... + 3³¹

3S - S = ( 3 + 3² + 3³ + 3⁴ + ... + 3³¹ ) - ( 1 + 3 + 3² + 3³ + ... + 3³⁰ )

2S = 3³¹ - 1

S = \(\frac{3^{31}-1}{2}\)

Answer 2

\(S=1+3+3^2+3^3+...+3^{30}\)

\(S=1+3\left(1+3^2+...+3^{29}\right)\)

\(S=1+3\left(S-3^{30}\right)\)

\(S=1+3S-3^{31}\)

\(2S=3^{31}-1\)

\(S=\frac{3^{31}-1}{2}\)

\(N=1+4+4^2+...+4^{132}=1+4\left(1+4^2+...+4^{131}\right)\)

\(N=1+3\left(N-4^{132}\right)\)

\(N=1+3N-4^{133}=\frac{4^{133}-1}{2}\)

Answer 3

 bn giải cả phần n nữa Huy Hoàng indonaca