Question

cho  P= 1+3^2+3^4+3^6+3^8+....+3^100. Và Q=(9^17/2)^3. so sánh P và Q

Answer 1

P = 1 + 3² + 3⁴ + 3⁶+......+3¹⁰⁰

3² P= 3²(1 + 3² + 3⁴ + 3⁶+......+3¹⁰⁰)

3²P= 3² + 3⁴ + 3⁶+......+3¹⁰⁰+3¹⁰²

3²P= (3² + 3⁴ + 3⁶+......+3¹⁰⁰+3¹⁰²)- (1 + 3² + 3⁴ + 3⁶+......+3¹⁰⁰ )

3² P= 3¹⁰² - 1

P= (3¹⁰² -1) :9

Q = (9¹⁷)³ / 2³

Q = 9⁵¹ / 8

Q = (3²)⁵¹ /8

Q = 3¹⁰² /8

Q= 3¹⁰² :8

=> P > Q

Vậy...

K chắc nha b

Answer 2

xét P=1+3^2+3^4+3^6+3^8+....+3^100

=> 3^2.P=3^2+3^4+3^6+3^8+3^10+...+3^102

9.P-P=(3^2+3^4+3^6+3^8+3^10+...+3^102)-(1+3^2+3^4+3^6+3^8+....+3^100)

8P=3^102-1

P=\(\frac{3^{102}-1}{8}\)

Xét Q :

\(\left(\frac{9^{17}}{2}\right)^3=\left[\frac{\left(3^2\right)^{17}}{2}\right]^3=\frac{\left(3^{34}\right)^3}{8}=\frac{3^{102}}{8}\)

mà 3^102-1<3^102

=>P<Q