S = (-3)⁰ + (-3)¹ + (-3)² + ... +  (-3)²⁰¹⁵

=> 3S = (-3)¹ + (-3)² + (-3)³ + ... +  (-3)²⁰¹⁶

=> 3S + S = [(-3)¹ + (-3)² + ... +  (-3)²⁰¹⁶] + [(-3)⁰ + (-3)¹ + ... +  (-3)²⁰¹⁵]

=> 4S = (-3)²⁰¹⁶ + (-3)⁰

=> S = \(\frac{\left(-3\right)^{2016}+\left(-3\right)^0}{4}\)

viết lại đề cho rõ phân số đi bn

S=1-3+3²-3³+...+3²⁰¹⁴-3²⁰¹⁵

=>3S=3-3²+...+3²⁰¹⁵-3²⁰¹⁶

=>3S+S=4S=(3-3²+...+3²⁰¹⁵-3²⁰¹⁶)+(1-3+...+3²⁰¹⁴-3²⁰¹⁵)

=>4S=-3²⁰¹⁶+1

=>S=\(-\frac{3^{2016}-1}{4}\)

\(S=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+\left(-3\right)^3+........+\left(-3\right)^{2015}\)

\(\Rightarrow-3S=\left(-3\right)^1+\left(-3\right)^2+\left(-3\right)^3+\left(-3\right)^4+......+\left(-3\right)^{2016}\)

\(\Rightarrow-4S=\left[\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{2016}\right]-\left[\left(-3\right)^0+\left(-3\right)^1+...+\left(-3\right)^{2015}\right]\)

\(\Rightarrow-4S=\left(-3\right)^{2016}-\left(-3\right)^0\Rightarrow-4S=3^{2016}-1\Rightarrow S=\frac{3^{2016}-1}{-4}\)

Trả lời:

\(S=\) \(\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\)\(\left(-3\right)^{2015}\)

\(-3S=\)\(\left(-3\right)^1+\left(-3\right)^2+...+\)\(\left(-3\right)^{2016}\)

\(-3S-S=\)\([\left(-3\right)^1+\left(-3\right)^2+...+\)\(\left(-3\right)^{2016}\)\(]\)\(-\)\([\)\(\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\)\(\left(-3\right)^{2015}\)\(]\)

\(\left(-3-1\right)S=\)\(\left(-3\right)^1+\left(-3\right)^2+...+\)\(\left(-3\right)^{2016}\)\(-\)\(\left(-3\right)^0-\left(-3\right)^1-\left(-3\right)^2-...-\)\(\left(-3\right)^{2015}\)

\(-4S=\)\(\left[\left(-3\right)^1-\left(-3\right)^1\right]\)\(+\)\(\left[\left(-3\right)^2-\left(-3\right)^2\right]\)\(+\)\(...\)\(+\)\(\left[\left(-3\right)^{2015}-\left(-3\right)^{2015}\right]\)\(+\)\(\left[\left(-3\right)^{2016}-\left(-3\right)^0\right]\)

\(-4S=\)\(0+0+...+0+\left(-3\right)^{2016}-1\)

\(-4S=\)\(3^{2016}-1\)

\(S=\frac{-3^{2016}+1}{4}\)

Vậy \(S=\frac{-3^{2016}+1}{4}\)

P/s: Không chắc có đúng ko. 

Hok tốt!

Vuong Dong Yet

Áp dụng công thức:

1 + 2³ + 3³ + ... + n³ = (1 + 2 + 3 + ... + n)² ta có

A = 1 + 2³ + 3³ + ... + 2015³ = (1 + 2 + 3 + ... + 2015)²

A = [(2015+1).2015:2]²

A = ( \(\dfrac{2016.2015}{2}\))²

A = (1008. 2015)²

A = 2031120²

( - 3) . S = (- 3) + ( - 3)² +...+ ( - 3)²⁰¹⁶

( - 3) . S - S = ( - 3)²⁰¹⁶ - 1

( - 4 ) S = ( - 3)²⁰¹⁶  - 1

S = [ ( - 3 )²⁰¹⁶ - 1 ] : ( - 4)

giai tat the

Ta có :

\(S=2015+\frac{2015}{1+2}+\frac{2015}{1+2+3}+...+\frac{2015}{1+2+3+..+2016}\)

    \(=2015.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+2016}\right)\)

    \(=2015.\left(1+\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(2016+1\right).2016}{2}}\right)\)

    \(=2015.\left(\frac{2}{2}+\frac{2}{2.\left(2+1\right)}+\frac{2}{3.\left(3+1\right)}+...+\frac{2}{2016.\left(2016+1\right)}\right)\)

    \(=2015.2.\left(\frac{1}{2}+\frac{1}{2.\left(2+1\right)}+\frac{1}{3.\left(3+1\right)}+...+\frac{1}{2016.\left(2016+1\right)}\right)\)

    \(=2015.2.\left(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)

    \(=2015.2.\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\) 

    \(=2015.2.\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{2017}\right)\)

    \(=2015.2.\left(1-\frac{1}{2017}\right)\)

    \(=2015.2.\frac{2016}{2017}\)

    =\(\frac{2015.2.2016}{2017}\)

    =\(\frac{8124480}{2017}\)

Vậy \(S=\frac{8124480}{2017}\)

Sai vì ngoài học tập ra còn cần phải siêng năng chăm chỉ trong các lĩnh vực khác nửa như giúp đỡ mọi người ,tham gia các hoạt động rèn luyện