Đặt S = 2.2² + 3.2³ + 4.2⁴ + ... + (n - 1).2^{n - 1} + n.2ⁿ 

<=> S = 2S - S = (2.2³ + 3.2⁴ +  4.2⁵ + .... + (n - 1).2ⁿ + n. 2^{n + 1}) - (2.2² + 3.2³ + 4.2⁴ + ... + (n - 1).2^{n - 1} + n.2ⁿ)

                S = (2.2³ - 3.2³) + (3.2⁴ - 4.2⁴) + (4.2⁵ - 5.2⁵) + .... + [(n - 1).2ⁿ - n.2ⁿ] + n.2^{n + 1} - 2.2²

                   = -(2³ + 2⁴ + 2⁵ + ... + 2ⁿ) + n.2^{n + 1} - 8

Đặt A = 2³ + 2⁴ + 2⁵ + ... + 2ⁿ

  <=> 2A - A = (2⁴ + 2⁵ + 2⁶ + ... + 2^{n + 1}) - (2³ + 2⁴ + 2⁵ + ... + 2ⁿ) 

  <=> A = 2^{n + 1} - 2³ 

Khi đó S = - 2^{n - 1} + 2³ + n.2^{n - 1} - 8

              = 2^{n - 1}.(n - 1) = 2^{n + 34}

         => n - 1 = 2^{n + 34} : 2^{n - 1}

          => n - 1 = 2^{n + 34 - n + 1}

          => n - 1 = 2³⁵

          => n = 2³⁵ + 1

N=34359738369 nha

Cường xo tính lại kết quả của Xyz mà cũng làm

làm giúp mình nha mai mình nộp r

1:

\(K=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot2^n-3^n}{2^{n+1}+3^{n+1}}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot2^n-3^n}{2^n\cdot2+3^n\cdot3}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot\dfrac{2^n}{3^n}-1}{\left(\dfrac{2}{3}\right)^n\cdot2+3}\)

\(=-\dfrac{1}{3}\)

2: 

\(\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^{n+1}}{3^{n+2}+4^n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{\left(\dfrac{3}{4}\right)^n-4}{\left(\dfrac{3}{4}\right)^n\cdot9+1}=-\dfrac{4}{1}=-4\)

a) \(A=2^{n-1}+2.2^{n+3}-8.2^{n-4}-16.2^n\)

\(=2^{n-1}+2^{n+3+1}-2^{n-4+3}-2^{n+4}\)

\(=2^{n-1}+2^{n+4}-2^{n-1}-2^{n+4}\)

\(=0\)

b) \(B=\left(3^{n+1}-2.2^n\right)\left(3^{n+1}+2.2^n\right)-3^{2n+2}+\left(8.2^{n-2}\right)^2\)

\(=\left(3^{n+1}-2^{n+1}\right)\left(3^{n+1}-2^{n+1}\right)-3^{2n+2}+2^{2n+2}\)

\(=3^{2n+2}-2^{2n+2}-3^{2n+2}+2^{2n+2}\)

\(=0\)