$#Leam$
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 ^^
