Isbot. Bu formulani isbotlash uchun bizga kerakli formulalarni isbotlab olaylik! Demak bizga quyidagi formulalar kerak!

Sana	16.11.2020
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Bog'liq
Tengelov Jasurbek Ajoyib formula isboti

Isbotlang:

𝟏

𝟒

+ 𝟐

𝟒

+ 𝟑

𝟒

+ ⋯ + 𝒏

𝟒

𝒏(𝒏 + 𝟏)(𝟐𝒏 + 𝟏)(𝟑𝒏

𝟐

+ 𝟑𝒏 − 𝟏)

𝟑𝟎

Isbot.

Bu formulani isbotlash uchun bizga kerakli formulalarni isbotlab olaylik!

Demak bizga quyidagi formulalar kerak!

𝟏. 𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏 =

𝒏(𝒏 + 𝟏)

𝟐

𝟐. 𝟏

𝟐

+ 𝟐

𝟐

+ 𝟑

𝟐

+ ⋯ + 𝒏

𝟐

𝒏(𝒏 + 𝟏)(𝟐𝒏 + 𝟏)

𝟔

𝟑. 𝟏

𝟑

+ 𝟐

𝟑

+ 𝟑

𝟑

+ ⋯ + 𝒏

𝟑

= (

𝒏(𝒏 + 𝟏)

𝟐

)

𝟐

1-formula isboti

(𝒏 + 𝟏)

𝟐

− 𝒏

𝟐

= 𝟐𝒏 + 𝟏

{

𝒏 = 𝟏 => 𝟐

𝟐

− 𝟏

𝟐

= 𝟐 ∙ 𝟏 + 𝟏

𝒏 = 𝟐 => 𝟑

𝟐

− 𝟐

𝟐

= 𝟐 ∙ 𝟐 + 𝟏

𝒏 = 𝟑 => 𝟒

𝟐

− 𝟑

𝟐

= 𝟐 ∙ 𝟑 + 𝟏

…

𝒏 = 𝒏 => (𝒏 + 𝟏)

𝟐

− 𝒏

𝟐

= 𝟐𝒏 + 𝟏

=> (𝒏 + 𝟏)

𝟐

− 𝟏

𝟐

= 𝟐 ∙ (𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏) + 𝒏 ∙ 𝟏

𝒏

𝟐

+ 𝟐𝒏 + 𝟏 − 𝟏 − 𝒏 = 𝟐 ∙ (𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏) => 𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏 =

𝒏(𝒏 + 𝟏)

𝟐

2-formula isboti

(𝒏 + 𝟏)

𝟑

− 𝒏

𝟑

= 𝟑𝒏

𝟐

+ 𝟑𝒏 + 𝟏

{

𝒏 = 𝟏 => 𝟐

𝟑

− 𝟏

𝟑

= 𝟑 ∙ 𝟏

𝟐

+ 𝟑 ∙ 𝟏 + 𝟏

𝒏 = 𝟐 => 𝟑

𝟑

− 𝟐

𝟑

= 𝟑 ∙ 𝟐

𝟐

+ 𝟑 ∙ 𝟐 + 𝟏

𝒏 = 𝟑 => 𝟒

𝟑

− 𝟑

𝟑

= 𝟑 ∙ 𝟑

𝟐

+ 𝟑 ∙ 𝟑 + 𝟏

…

𝒏 = 𝒏 => (𝒏 + 𝟏)

𝟑

− 𝒏

𝟑

= 𝟑𝒏

𝟐

+ 𝟑𝒏 + 𝟏

=> (𝒏 + 𝟏)

𝟑

− 𝟏

𝟑

= 𝟑 ∙ (𝟏

𝟐

+ 𝟐

𝟐

+ 𝟑

𝟐

+ ⋯ + 𝒏

𝟐

) + 𝟑 ∙ (𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏) + 𝒏

𝒏

𝟑

+ 𝟑𝒏

𝟐

+ 𝟑𝒏 − 𝟑 ∙

𝒏(𝒏 + 𝟏)

𝟐

− 𝒏 = 𝟑 ∙ (𝟏

𝟐

+ 𝟐

𝟐

+ 𝟑

𝟐

+ ⋯ + 𝒏

𝟐

) 𝒔𝒐𝒅𝒅𝒂𝒍𝒂𝒔𝒉𝒕𝒊𝒓𝒂𝒎𝒊𝒛

𝟐𝒏

𝟑

+ 𝟑𝒏

𝟐

+ 𝒏

𝟐

= 𝟑 ∙ (𝟏

𝟐

+ 𝟐

𝟐

+ 𝟑

𝟐

+ ⋯ + 𝒏

𝟐

) => (𝟏

𝟐

+ 𝟐

𝟐

+ 𝟑

𝟐

+ ⋯ + 𝒏

𝟐

) =

𝒏(𝒏 + 𝟏)(𝟐𝒏 + 𝟏)

𝟔

3-formula isboti

(𝒏 + 𝟏)

𝟒

− 𝒏

𝟒

= 𝟒𝒏

𝟑

+ 𝟔𝒏

𝟐

+ 𝟒𝒏 + 𝟏

{

𝒏 = 𝟏 => 𝟐

𝟒

− 𝟏

𝟒

= 𝟒 ∙ 𝟏

𝟑

+ 𝟔 ∙ 𝟏

𝟐

+ 𝟒 ∙ 𝟏 + 𝟏

𝒏 = 𝟐 => 𝟑

𝟒

− 𝟐

𝟒

= 𝟒 ∙ 𝟐

𝟑

+ 𝟔 ∙ 𝟐

𝟐

+ 𝟒 ∙ 𝟐 + 𝟏

𝒏 = 𝟑 => 𝟒

𝟒

− 𝟑

𝟒

= 𝟒 ∙ 𝟑

𝟑

+ 𝟔 ∙ 𝟑

𝟐

+ 𝟒 ∙ 𝟑 + 𝟏

…

𝒏 = 𝒏 => (𝒏 + 𝟏)

𝟒

− 𝒏

𝟒

= 𝟒𝒏

𝟑

+ 𝟔𝒏

𝟐

+ 𝟒𝒏 + 𝟏

=> (𝒏 + 𝟏)

𝟒

− 𝟏

𝟒

= 𝟒(𝟏

𝟑

+ 𝟐

𝟑

+ 𝟑

𝟑

+ ⋯ + 𝒏

𝟑

) + 𝟔(𝟏

𝟐

+ 𝟐

𝟐

+ 𝟑

𝟐

+ ⋯ + 𝒏

𝟐

) +

+𝟒(𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏) + 𝒏 =>

(𝒏 + 𝟏)

𝟒

− 𝟏 = 𝟒 ∙ (𝟏

𝟑

+ 𝟐

𝟑

+ 𝟑

𝟑

+ ⋯ + 𝒏

𝟑

) + 𝟔 ∙

𝒏(𝒏 + 𝟏)(𝟐𝒏 + 𝟏)

𝟔

+ 𝟒 ∙

𝒏(𝒏 + 𝟏)

𝟐

+ 𝒏 =>

=> (𝒏 + 𝟏)

𝟒

− 𝒏(𝒏 + 𝟏)(𝟐𝒏 + 𝟏) − 𝟐𝒏(𝒏 + 𝟏) − (𝒏 + 𝟏) = 𝟒 ∙ (𝟏

𝟑

+ 𝟐

𝟑

+ 𝟑

𝟑

+ ⋯ + 𝒏

𝟑

) =>

𝑺𝒐𝒅𝒅𝒂𝒍𝒂𝒔𝒉𝒕𝒊𝒓𝒂𝒎𝒊𝒛 => (𝟏

𝟑

+ 𝟐

𝟑

+ 𝟑

𝟑

+ ⋯ + 𝒏

𝟑

) = (

𝒏(𝒏 + 𝟏)

𝟐

)

𝟐

Endi isbotlashimiz kerak bo’lgan formulaga qaytamiz!

(𝒏 + 𝟏)

𝟓

− 𝒏

𝟓

= 𝟓𝒏

𝟒

+ 𝟏𝟎𝒏

𝟑

+ 𝟏𝟎𝒏

𝟐

+ 𝟓𝒏 + 𝟏

{

𝒏 = 𝟏 => 𝟐

𝟓

− 𝟏

𝟓

= 𝟓 ∙ 𝟏

𝟒

+ 𝟏𝟎 ∙ 𝟏

𝟑

+ 𝟏𝟎 ∙ 𝟏

𝟐

+ 𝟓 ∙ 𝟏 + 𝟏

𝒏 = 𝟐 => 𝟑

𝟓

− 𝟐

𝟓

= 𝟓 ∙ 𝟐

𝟒

+ 𝟏𝟎 ∙ 𝟐

𝟑

+ 𝟏𝟎 ∙ 𝟐

𝟐

+ 𝟓 ∙ 𝟐 + 𝟏

𝒏 = 𝟑 => 𝟒

𝟓

− 𝟑

𝟓

= 𝟓 ∙ 𝟑

𝟒

+ 𝟏𝟎 ∙ 𝟑

𝟑

+ 𝟏𝟎 ∙ 𝟑

𝟐

+ 𝟓 ∙ 𝟏 + 𝟏

…

𝒏 = 𝒏 => (𝒏 + 𝟏)

𝟓

− 𝒏

𝟓

= 𝟓𝒏

𝟒

+ 𝟏𝟎𝒏

𝟑

+ 𝟏𝟎𝒏

𝟐

+ 𝟓𝒏 + 𝟏

=> (𝒏 + 𝟏)

𝟓

− 𝟏

𝟓

= 𝟓 ∙ (𝟏

𝟒

+ 𝟐

𝟒

+ 𝟑

𝟒

+ ⋯ + 𝒏

𝟒

) + 𝟏𝟎 ∙ (𝟏

𝟑

+ 𝟐

𝟑

+ 𝟑

𝟑

+ ⋯ + 𝒏

𝟑

) +

+𝟏𝟎 ∙ (𝟏

𝟐

+ 𝟐

𝟐

+ 𝟑

𝟐

+ ⋯ + 𝒏

𝟐

) + 𝟓 ∙ (𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏) + 𝒏

𝟓 ∙ (𝟏

𝟒

+ 𝟐

𝟒

+ 𝟑

𝟒

+ ⋯ + 𝒏

𝟒

) = (𝒏 + 𝟏)

𝟓

− 𝟏𝟎 ∙ (

𝒏(𝒏 + 𝟏)

𝟐

)

𝟐

− 𝟏𝟎 ∙

𝒏(𝒏 + 𝟏)(𝟐𝒏 + 𝟏)

𝟔

−

−𝟓 ∙

𝒏(𝒏 + 𝟏)

𝟐

− (𝒏 + 𝟏) 𝒆𝒏𝒅𝒊 𝒐

′

𝒏𝒈 𝒕𝒐𝒎𝒐𝒏𝒏𝒊 𝒔𝒐𝒅𝒅𝒂𝒍𝒂𝒔𝒉𝒕𝒊𝒓𝒂𝒎𝒊𝒛

(𝒏 + 𝟏) ((𝒏 + 𝟏)

𝟒

−

𝟓 ∙ 𝒏

𝟐

∙ (𝒏 + 𝟏)

𝟐

−

𝟓 ∙ 𝒏 ∙ (𝟐𝒏 + 𝟏)

𝟑

−

𝟓𝒏

𝟐

− 𝟏) 𝒔𝒐𝒅𝒅𝒂𝒍𝒂𝒔𝒉𝒕𝒊𝒓𝒔𝒂𝒌

𝟓 ∙ (𝟏

𝟒

+ 𝟐

𝟒

+ 𝟑

𝟒

+ ⋯ + 𝒏

𝟒

) =

𝒏(𝒏 + 𝟏)(𝟐𝒏 + 𝟏)(𝟑𝒏

𝟐

+ 𝟑𝒏 − 𝟏)

𝟔

(𝟏

𝟒

+ 𝟐

𝟒

+ 𝟑

𝟒

+ ⋯ + 𝒏

𝟒

) =

𝒏(𝒏 + 𝟏)(𝟐𝒏 + 𝟏)(𝟑𝒏

𝟐

+ 𝟑𝒏 − 𝟏)

𝟑𝟎

𝒌𝒆𝒍𝒊𝒃 𝒄𝒉𝒊𝒒𝒂𝒅𝒊.

Yechim muallifi: Qoraqolpog’iston Respublikasi Qo’ng’irot tumani 35-maktab

10-sinf o’quvchisi Tengelov Jasurbek

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