Question

1+2×3+3×7+...+n×(2^n-1) find sum.​

Answer 1

Заметим что

$1 + 2\cdot(1+2+3+...+2^{n-1}) = 1+2+3+...+2^{n}\\1+(1+2+3+...+2^{n-1}) = 2^n\\2^{n-1} =1+2+3+...+2^{n}-1$

Поэтому наша сумма

$S_n = \\1\cdot1+\\2\cdot(1+2)+\\3\cdot(1+2+4)+\\4\cdot(1+2+4+8)+\\\...\\n\cdot(1+2+4+8+...+2^{n -1})$

$2S_n = \\1\cdot2+\\2\cdot(2+4)+\\3\cdot(2+4+8)+\\...\\n\cdot(2+4+8+...+2^n)$

В то же время если мы прибавим к первой строчке 1, ко второй 2 к третьей 3 и так далее мы в итоге прибавим $n(n+1)/2$ и получим

$2S_n +n(n+1)/2= \\1\cdot(1+2)+\\2\cdot(1+2+4)+\\3\cdot(1+2+4+8)+\\...\\n\cdot(1+2+4+8+...+2^n)$

Теперь прибавим $1=2^1-1$, потом к первой строчке прибавим $1+2=2^2-1$, к третьей строчке $1+2+4=2^3-1$ и так далее, в итоге мы прибавим $2+2^2+2^2+...2^n+2^{n+1}-(n+1)=2^{n+2}-(n+3)$ и получим

$2S_n +n(n+1)/2+2^{n+2}-n-3= \\1+\\2\cdot(1+2)+\\3\cdot(1+2+4)+\\4\cdot(1+2+4+8)+\\...\\(n+1)\cdot(1+2+4+8+...+2^n) = S_{n+1}$

или короче

$\displaystyle 2S_n+\frac{n^2-n-6}{2}+2^{n+2} = S_{n+1}$

c другой стороны

$S_{n+1} = S_n+(n+1)(2^{n+1}-1)$

поэтому

$\displaystyle S_n+\frac{n^2-n-6}{2}+2^{n+2} = S_{n+1}-S_n = (n+1)(2^{n+1}-1)$

и окончательно

$\displaystyle S_n= (n+1)(2^{n+1}-1)-\frac{n^2-n-6}{2}-2^{n+2} = (n-1)\cdot2^{n+1}-\frac{n^2+n-4}{2}$

Answer 2

Ответ:

...................................................

Пошаговое объяснение:

Answer 3

To find the sum of the series 1 + 2×3 + 3×7 + ... + n×(2^n-1), we can use the formula for the sum of an arithmetic series. However, it seems that the given series does not follow a simple arithmetic pattern.

To better understand the series, let's break it down:

1. The first term is 1, which is obtained by multiplying 1 by (2^1 - 1). 2. The second term is 2×3, which is obtained by multiplying 2 by (2^2 - 1). 3. The third term is 3×7, which is obtained by multiplying 3 by (2^3 - 1). 4. And so on...

It appears that each term is obtained by multiplying the term number by (2^term number - 1).

Let's calculate the sum of the series for a specific value of n, say n = 4:

1 + 2×3 + 3×7 + 4×15

= 1×(2^1 - 1) + 2×(2^2 - 1) + 3×(2^3 - 1) + 4×(2^4 - 1)

= 1×(2 - 1) + 2×(4 - 1) + 3×(8 - 1) + 4×(16 - 1)

= 1 + 2×3 + 3×7 + 4×15

= 1 + 6 + 21 + 60

= 88

So, the sum of the series for n = 4 is 88.

To find the sum of the series for any value of n, we can use a loop or a recursive function to calculate each term and add them up. Here's an example code snippet in Python:

def calculate_sum(n):
    sum = 0
    for i in range(1, n+1):
        term = i * (2**i - 1)
        sum += term
    return sum

n = 4
series_sum = calculate_sum(n)
print("The sum of the series for n =", n, "is", series_sum)

Please note that the code snippet provided is just an example and may need to be modified based on the programming language you are using.

I hope this helps! Let me know if you have any further questions.