Tính $\dfrac{1+2^2+3^2+\ \!…\,+\,100^2}{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\ \!…\,+\,\dfrac{1}{100^2}}$. 24/07/2021 Bởi Eden Tính $\dfrac{1+2^2+3^2+\ \!…\,+\,100^2}{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\ \!…\,+\,\dfrac{1}{100^2}}$.
$#Leam$ Gọi tử là A ⇒ A = 1 + 2² + 3² + … + 100² = 1 + 2(1+1) + 3(2+1) + … + 100.(99+1) = 1 + 1.2 + 2 + 2.3 + 3 + … + 99.100 + 100 = ( 1.2 + 2.3 + … + 99.100) + ( 1+2+….+ 100) Gọi ( 1.2 + 2.3 + … + 99.100) là B ⇒ B = 1.2 + 2.3 + … + 99.100 3B = 1.2.3 + 2.3.(4-1) + … + 99.100.(101- 98 ) = 99.100.101 B = 99.100.101:3 = 333300 Gọi ( 1+2+….+ 100) là C ⇒ C = 1+2+….+ 100 Tính … ⇒ = 5050 ⇒ A = 333300 + 5050 = 338350 Gọi mẫu là D ⇒ D = 1 + $\dfrac{1}{2²}$ + $\dfrac{1}{3²}$ + … + $\dfrac{1}{100²}$ = $\dfrac{1}{1²}$ + $\dfrac{1}{2²}$ + $\dfrac{1}{3²}$ + … + $\dfrac{1}{100²}$ =$\dfrac{1}{1}$ + $\dfrac{1}{2.(1+1)}$ + $\dfrac{1}{3.(2+1)}$ + … + $\dfrac{1}{100.(99+1)}$ = $\dfrac{1}{1}$ + $\dfrac{1}{1.2}$ + $\dfrac{1}{2}$ + … + $\dfrac{1}{99.100}$ + $\dfrac{1}{100}$ = ( $\dfrac{1}{1}$ + $\dfrac{1}{2}$ + … + $\dfrac{1}{100}$ ) + ( $\dfrac{1}{1.2}$ + $\dfrac{1}{2.3}$ + … + $\dfrac{1}{99.100}$ ) Gọi $\dfrac{1}{1}$ + $\dfrac{1}{2}$ + … + $\dfrac{1}{100}$ là E ⇒ E = $\dfrac{1}{1}$ + $\dfrac{1}{2}$ + … + $\dfrac{1}{100}$ Tính … ( bn lên mạng tra chắc có vì mik lười vt ⇒ E = $\dfrac{199}{100}$ Gọi $\dfrac{1}{1.2}$ + $\dfrac{1}{2.3}$ + … + $\dfrac{1}{99.100}$ là F ⇒ F = $\dfrac{1}{1.2}$ + $\dfrac{1}{2.3}$ + … + $\dfrac{1}{99.100}$ = $\dfrac{1}{1}$ – $\dfrac{1}{2}$ + … + $\dfrac{1}{99}$ – $\dfrac{1}{100}$ = 1 – $\dfrac{1}{100}$ = $\dfrac{99}{100}$ ⇒ D = $\dfrac{199}{100}$ + $\dfrac{99}{100}$ = $\dfrac{298}{100}$ = $\dfrac{149}{50}$ Tổng $\dfrac{1 + 2² + 3² + … + 100²}{1 + $\dfrac{1}{2²}$ + $\dfrac{1}{3²}$ + … + $\dfrac{1}{100²}$}$ = $\dfrac{338350}{$\dfrac{149}{50}$}$ = $\dfrac{16917500}{149}$ Nếu có sai sót j mong bn bỏ qua vì mik hok toán hơi nu CHUCBANHOKTOT ^^ Bình luận
$#Leam$
Gọi tử là A
⇒ A = 1 + 2² + 3² + … + 100²
= 1 + 2(1+1) + 3(2+1) + … + 100.(99+1)
= 1 + 1.2 + 2 + 2.3 + 3 + … + 99.100 + 100
= ( 1.2 + 2.3 + … + 99.100) + ( 1+2+….+ 100)
Gọi ( 1.2 + 2.3 + … + 99.100) là B
⇒ B = 1.2 + 2.3 + … + 99.100
3B = 1.2.3 + 2.3.(4-1) + … + 99.100.(101- 98 )
= 99.100.101
B = 99.100.101:3
= 333300
Gọi ( 1+2+….+ 100) là C
⇒ C = 1+2+….+ 100
Tính …
⇒ = 5050
⇒ A = 333300 + 5050
= 338350
Gọi mẫu là D
⇒ D = 1 + $\dfrac{1}{2²}$ + $\dfrac{1}{3²}$ + … + $\dfrac{1}{100²}$
= $\dfrac{1}{1²}$ + $\dfrac{1}{2²}$ + $\dfrac{1}{3²}$ + … + $\dfrac{1}{100²}$
=$\dfrac{1}{1}$ + $\dfrac{1}{2.(1+1)}$ + $\dfrac{1}{3.(2+1)}$ + … + $\dfrac{1}{100.(99+1)}$
= $\dfrac{1}{1}$ + $\dfrac{1}{1.2}$ + $\dfrac{1}{2}$ + … + $\dfrac{1}{99.100}$ + $\dfrac{1}{100}$
= ( $\dfrac{1}{1}$ + $\dfrac{1}{2}$ + … + $\dfrac{1}{100}$ ) + ( $\dfrac{1}{1.2}$ + $\dfrac{1}{2.3}$ + … + $\dfrac{1}{99.100}$ )
Gọi $\dfrac{1}{1}$ + $\dfrac{1}{2}$ + … + $\dfrac{1}{100}$ là E
⇒ E = $\dfrac{1}{1}$ + $\dfrac{1}{2}$ + … + $\dfrac{1}{100}$
Tính … ( bn lên mạng tra chắc có vì mik lười vt
⇒ E = $\dfrac{199}{100}$
Gọi $\dfrac{1}{1.2}$ + $\dfrac{1}{2.3}$ + … + $\dfrac{1}{99.100}$ là F
⇒ F = $\dfrac{1}{1.2}$ + $\dfrac{1}{2.3}$ + … + $\dfrac{1}{99.100}$
= $\dfrac{1}{1}$ – $\dfrac{1}{2}$ + … + $\dfrac{1}{99}$ – $\dfrac{1}{100}$
= 1 – $\dfrac{1}{100}$
= $\dfrac{99}{100}$
⇒ D = $\dfrac{199}{100}$ + $\dfrac{99}{100}$
= $\dfrac{298}{100}$
= $\dfrac{149}{50}$
Tổng $\dfrac{1 + 2² + 3² + … + 100²}{1 + $\dfrac{1}{2²}$ + $\dfrac{1}{3²}$ + … + $\dfrac{1}{100²}$}$ = $\dfrac{338350}{$\dfrac{149}{50}$}$ = $\dfrac{16917500}{149}$
Nếu có sai sót j mong bn bỏ qua vì mik hok toán hơi nu
CHUCBANHOKTOT ^^