各位高手,教下我點樣解題......15分

Here, we also need to know the following formula

1^2 + 2^2 + … + n^2 = n(n+1)(2n+1)/6
1^3 + 2^3 + … + n^3 = [n(n+1)/2]^2


(x+1)^5 - x^5 = 5x^4 +10x^3 + 10x^2 + 5x + 1
∑ (from 1 to 99) [(x+1)^5 – x^5] = ∑ (from 1 to 99) [5x^4 +10x^3 + 10x^2 + 5x + 1]

∑ (from 2 to 100) [x^5] - ∑ (from 1 to 99) [x^5] = ∑ (from 1 to 99) 5x^4 + ∑ (from 1 to 99) 10x^3 + ∑ (from 1 to 99) 10x^2 + ∑ (from 1 to 99) 5x + ∑ (from 1 to 99) 1

100^5 - 1^5 = 5∑ (from 1 to 99) x^4 + 10∑ (from 1 to 99) x^3 + 10∑ (from 1 to 99) x^2 + 5∑ (from 1 to 99) x + 99

10^10 - 1 = 5∑ (from 1 to 99) x^4 + 10[(99)(99+1)/2]^2 + 10[(99)(99+1)(2*99+1)/6] + 5(99)(99+1)/2 + 99

5∑ (from 1 to 99) x^4 = 10^10 - 1 - 10[(99)(99+1)/2]^2 - 10[(99)(99+1)(2*99+1)/6] - 5(99)(99+1)/2 – 99
5∑ (from 1 to 99) x^4 = 9751666650
∑ (from 1 to 99) x^4 = 1950333330
So, 1^4+2^4+3^4+...+99^4 = 1950333330

我諗黎諗去都諗唔到條公式點幫手計個ans
最後我自己喪加
加出個答案為1,950,333,330

你用≡這符號，又沒有(mod xx)，是沒有可能的，你是否遺漏了題目的一部分呢？
你前用≡，後用=，叫人無法明白。

你用≡這符號，又沒有(mod xx)，是沒有可能的，你是否遺漏了題目的一部分呢？
你前用≡，後用=，叫人無法明白。