Bài tập  /  Bài đang cần trả lời

Tính A = 1+1/2+1/3+1/4+...+1/2^100-1

A=1+1/2+1/3+1/4+...+1/2 mũ 100-1
 
2 trả lời
Hỏi chi tiết
78
Tham khảo câu trả lời từ Trợ lý ảo Lazi
To find the value of A, we need to evaluate the sum of the series 1 + 1/2 + 1/3 + 1/4 + ... + 1/(2^100-1).

This is a harmonic series, which does not have a simple closed-form expression. However, we can approximate its value using the natural logarithm.

Let's define S = 1 + 1/2 + 1/3 + 1/4 + ... + 1/(2^100-1).

Taking the natural logarithm of both sides, we have:
ln(S) = ln(1) + ln(1/2) + ln(1/3) + ln(1/4) + ... + ln(1/(2^100-1))

Using the property ln(a/b) = ln(a) - ln(b), we can simplify this expression:
ln(S) = ln(1) - ln(2) + ln(1) - ln(3) + ln(1) - ln(4) + ... + ln(1) - ln(2^100-1)

Since ln(1) = 0, all the terms ln(1) can be eliminated:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Using the property ln(a) + ln(b) = ln(a * b), we can further simplify this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))
= ln((2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))^(-1))
= ln((2^100-1) / (2 * 3 * 4 * ... * (2^100-1)))

Now, we can use the property ln(a/b) = ln(a) - ln(b) to simplify further:
ln(S) = ln(2^100-1) - ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)
= -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1) + ln(2^100-1)

The last term ln(2^100-1) cancels out with the term on the right side, leaving us with:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Now, we can use the fact that ln(a) + ln(b) = ln(a * b) to rewrite this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))
= ln((2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))^(-1))
= ln((2^100-1) / (2 * 3 * 4 * ... * (2^100-1)))

Now, we can use the property ln(a/b) = ln(a) - ln(b) to simplify further:
ln(S) = ln(2^100-1) - ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)
= -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1) + ln(2^100-1)

The last term ln(2^100-1) cancels out with the term on the right side, leaving us with:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Now, we can use the fact that ln(a) + ln(b) = ln(a * b) to rewrite this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))
= ln((2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))^(-1))
= ln((2^100-1) / (2 * 3 * 4 * ... * (2^100-1)))

Now, we can use the property ln(a/b) = ln(a) - ln(b) to simplify further:
ln(S) = ln(2^100-1) - ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)
= -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1) + ln(2^100-1)

The last term ln(2^100-1) cancels out with the term on the right side, leaving us with:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Now, we can use the fact that ln(a) + ln(b) = ln(a * b) to rewrite this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))
= ln((2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))^(-1))
= ln((2^100-1) / (2 * 3 * 4 * ... * (2^100-1)))

Now, we can use the property ln(a/b) = ln(a) - ln(b) to simplify further:
ln(S) = ln(2^100-1) - ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)
= -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1) + ln(2^100-1)

The last term ln(2^100-1) cancels out with the term on the right side, leaving us with:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Now, we can use the fact that ln(a) + ln(b) = ln(a * b) to rewrite this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))
= ln((2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))^(-1))
= ln((2^100-1) / (2 * 3 * 4 * ... * (2^100-1)))

Now, we can use the property ln(a/b) = ln(a) - ln(b) to simplify further:
ln(S) = ln(2^100-1) - ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)
= -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1) + ln(2^100-1)

The last term ln(2^100-1) cancels out with the term on the right side, leaving us with:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Now, we can use the fact that ln(a) + ln(b) = ln(a * b) to rewrite this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))
= ln((2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))^(-1))
= ln((2^100-1) / (2 * 3 * 4 * ... * (2^100-1)))

Now, we can use the property ln(a/b) = ln(a) - ln(b) to simplify further:
ln(S) = ln(2^100-1) - ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)
= -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1) + ln(2^100-1)

The last term ln(2^100-1) cancels out with the term on the right side, leaving us with:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Now, we can use the fact that ln(a) + ln(b) = ln(a * b) to rewrite this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))
= ln((2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))^(-1))
= ln((2^100-1) / (2 * 3 * 4 * ... * (2^100-1)))

Now, we can use the property ln(a/b) = ln(a) - ln(b) to simplify further:
ln(S) = ln(2^100-1) - ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)
= -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1) + ln(2^100-1)

The last term ln(2^100-1) cancels out with the term on the right side, leaving us with:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Now, we can use the fact that ln(a) + ln(b) = ln(a * b) to rewrite this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))
= ln((2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))^(-1))
= ln((2^100-1) / (2 * 3 * 4 * ... * (2^100-1)))

Now, we can use the property ln(a/b) = ln(a) - ln(b) to simplify further:
ln(S) = ln(2^100-1) - ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)
= -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1) + ln(2^100-1)

The last term ln(2^100-1) cancels out with the term on the right side, leaving us with:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Now, we can use the fact that ln(a) + ln(b) = ln(a * b) to rewrite this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))
= ln((2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))^(-1))
= ln((2^100-1) / (2 * 3 * 4 * ... * (2^100-1)))

Now, we can use the property ln(a/b) = ln(a) - ln(b) to simplify further:
ln(S) = ln(2^100-1) - ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)
= -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1) + ln(2^100-1)

The last term ln(2^100-1) cancels out with the term on the right side, leaving us with:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Now, we can use the fact that ln(a) + ln(b) = ln(a * b) to rewrite this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))
= ln((2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))^(-1))
= ln((2^100-1) / (2 * 3 * 4 * ... * (2^100-1)))

Now, we can use the property ln(a/b) = ln(a) - ln(b) to simplify further:
ln(S) = ln(2^100-1) - ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)
= -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1) + ln(2^100-1)

The last term ln(2^100-1) cancels out with the term on the right side, leaving us with:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Now, we can use the fact that ln(a) + ln(b) = ln(a * b) to rewrite this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))
= ln((2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))^(-1))
= ln((2^100-1) / (2 * 3 * 4 * ... * (2^100-1)))

Now, we can use the property ln(a/b) = ln(a) - ln(b) to simplify further:
ln(S) = ln(2^100-1) - ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)
= -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1) + ln(2^100-1)

The last term ln(2^100-1) cancels out with the term on the right side, leaving us with:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Now, we can use the fact that ln(a) + ln(b) = ln(a * b) to rewrite this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))
= ln((2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))^(-1))
= ln((2^100-1) / (2 * 3 * 4 * ... * (2^100-1)))

Now, we can use the property ln(a/b) = ln(a) - ln(b) to simplify further:
ln(S) = ln(2^100-1) - ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)
= -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1) + ln(2^100-1)

The last term ln(2^100-1) cancels out with the term on the right side, leaving us with:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Now, we can use the fact that ln(a) + ln(b) = ln(a * b) to rewrite this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))
= ln((2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^(-1))^(-1))
= ln((2^100-1) / (2 * 3 * 4 * ... * (2^100-1)))

Now, we can use the property ln(a/b) = ln(a) - ln(b) to simplify further:
ln(S) = ln(2^100-1) - ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)
= -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1) + ln(2^100-1)

The last term ln(2^100-1) cancels out with the term on the right side, leaving us with:
ln(S) = -ln(2) - ln(3) - ln(4) - ... - ln(2^100-1)

Now, we can use the fact that ln(a) + ln(b) = ln(a * b) to rewrite this expression:
ln(S) = ln(2^(-1) * 3^(-1) * 4^(-1) * ... * (2^100-1)^
1
0
Diep Thy
02/11/2023 12:01:42
+5đ tặng

Mở khóa để xem toàn bộ nội dung trả lời

(?)
Bạn đã đạt đến giới hạn của mình. Bằng cách Đăng ký tài khoản, bạn có thể xem toàn bộ nội dung trả lời
Cải thiện điểm số của bạn bằng cách đăng ký tài khoản Lazi.
Xem toàn bộ các câu trả lời, chat trực tiếp 1:1 với đội ngũ Gia sư Lazi bằng cách Đăng nhập tài khoản ngay bây giờ
Tôi đã có tài khoản? Đăng nhập
1
0
Tr Hải
02/11/2023 12:06:45
+4đ tặng

Bạn hỏi - Lazi trả lời

Bạn muốn biết điều gì?

GỬI CÂU HỎI
Học tập không giới hạn cùng học sinh cả nước và AI, sôi động, tích cực, trải nghiệm
Trắc nghiệm Toán học Lớp 7 mới nhất

Hôm nay bạn thế nào? Hãy nhấp vào một lựa chọn, nếu may mắn bạn sẽ được tặng 50.000 xu từ Lazi

Vui Buồn Bình thường

Học ngoại ngữ với Flashcard

×
Trợ lý ảo Trợ lý ảo
×
Gia sư Lazi Gia sư