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1) Số số hạng là: \(\frac{2x-1-1}{2}+1=\frac{2x-2}{2}+1=\frac{2\left(x-1\right)}{2}+1=x-1+1=x\)
Tổng là \(\frac{\left(1+2x-1\right).x}{2}=225\)
\(\frac{2.x^2}{2}=225\)
x2=225
x=15
Đợi chút mình làm câu b. Mỏi tay quá
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A = 3 + 32 + 33 + ... + 3100
3A = 32 + 33 + 34 + ... + 3101
3A - A = 3101 - 3
2A = 3101 - 3
Ta có:
2A + 3 = 3n
3101 - 3 +3 = 3n
3101 = 3n
=> n = 101
Vậy n = 101
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A = 3+32+33+...+3100
3A = 32+33+34+...+3101
2A = 3A - A = 3101 - 3
=> 2A + 3 = 3101
Theo đề bài: 2A + 3 = 3n
=> 3101 = 3n
=> n = 101
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3A = 3^2+3^3+...+3^101
3A-A=3^2+3^3+...+3^101-(3+3^2+3^3+...+3^100)
2A=3^101-A
2A+A=3n
Suy ra : 3^101-3+3=3n
Suy ra : 3^101=3n
Suy ra : n=3^100
bấm đúng cho mik vs nha
\(A=3+3^2+...+3^{100}\)
\(3A=3^2+3^3+....+3^{101}\)
\(3A-A=\left(3^2-3^2\right)+....+\left(3^{100}-3^{100}\right)+3^{101}-3\)
2A = 3101 - 3
\(A=\frac{3^{101}-3}{2}\)
2A + 3 = \(\frac{3^{101}-3}{2}.2+3=3^{101}-3+3=3^{101}\)
Vậy n = 101
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A = 3 + 32 + 33 + ... + 3100
3A = 32 + 33 + 34 + ... + 3101
3A - A = 3101 - 3
2A = 3101 - 3
Ta có:
2A + 3 = 3n
3101 - 3 +3 = 3n
3101 = 3n
=> n = 101
Vậy n = 101
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3A=32+33+......+3101
3A-A=3101-3
A=3101-2:2
2A+3=3n
2x3101-3:2+3=3n
3101-3+3=3n
3101=3n
n=101
3A=32+33+......+3101
3A-A=3101-3
A=3101-2:2
2A+3=3n
2x3101-3:2+3=3n
3101-3+3=3n
3101=3n
n=101
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A=3+3^2+3^3+..........+3^99+3^100
3A=3^2+3^3+...............+3^100+3^101
=> 3A-A= (3^2+3^3+......+3^100+3^101) - (3+3^2+3^3+........+3^99+3^100)
=> 2A= 3^101 - 3
=>2A+3=3^101
=>3^n=3^101
=> n=101
Ta có:
\(A=3+3^2+3^3+...+3^{99}+3^{100}\)
\(2A=3^2+3^3+3^4+...+3^{100}+3^{101}\)
\(2A-A=\left(3^2+3^3+3^4+...+3^{100}+3^{101}\right)-\left(3+3^2+3^3+...+3^{99}+3^{100}\right)\)\(A=3^{101}-3\)
\(2A+3=3^{101}-3+3=3^{101}=3^n\)
\(n=101\)
A=3+32+33+...+3100
3A=32+33+34+....+3101
3A-A=3101-3
2A=3101-3
b) 2A+3=3101
mà 2A+3=3x
nên 3x=3101
-> x=101
x=101 . Ai k mình mình sẽ lại cho