Last of Area Bonanza

Calculus Level 4

Let ∣ x ∣ < 1 |x| < 1 .

If f ( x ) = ∑ n = 1 ∞ n 5 x n f(x) = \sum_{n = 1}^{\infty} n^5 x^n and g ( x ) = ∑ n = 1 ∞ n 4 x n g(x) = \sum_{n = 1}^{\infty} n^4 x^n and the area A A of the region bounded by f f and g g on [ − 1 4 , 0 ] [\dfrac{-1}{4},0] can be expressed as A = c b b − a ln ⁡ ( b a a ) A = \dfrac{c}{b^b} - a\ln(\dfrac{b}{a^a}) , where a , b a,b and c c are coprime positive integers, find a + b + c a + b + c .


The answer is 1436.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
May 10, 2018

Let ∣ x ∣ < 1 |x| < 1 .

∑ n = 1 ∞ n x n = ∑ j = 1 ∞ ∑ n = j ∞ x n = 1 1 − x ∑ j = 1 ∞ x j = 1 1 − x ( x ) 1 1 − x = x ( 1 − x ) 2 \sum_{n = 1}^{\infty} n x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} x^n = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} x^{j} = \dfrac{1}{1 - x}(x)\dfrac{1}{1 - x} = \dfrac{x}{(1 - x)^2}

( x x − 1 ) ( 1 x ) ( x x − 1 ) = x ( x − 1 ) 2 . (\dfrac{x}{x - 1})(\dfrac{1}{x})(\dfrac{x}{x - 1}) = \dfrac{x}{(x - 1)^2}.

∑ n = 1 ∞ n 2 x n = ∑ j = 1 ∞ ( ∑ n = j ∞ n x n ) = \sum_{n = 1}^{\infty} n^2 x^n = \sum_{j = 1}^{\infty} (\sum_{n = j}^{\infty} n x^n) = 1 1 − x ∑ j = 1 ∞ ( j x j ) + x ( 1 − x ) 2 ∑ j = 1 ∞ x j = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} (j x^j) + \dfrac{x}{(1 - x)^2}\sum_{j = 1}^{\infty} x^j =

( 1 1 − x ) ( x ( 1 − x ) 2 ) + ( x ( 1 − x ) 2 ) ( x 1 − x ) = (\dfrac{1}{1 - x})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x}{(1 - x)^2})(\dfrac{x}{1 - x}) = x 2 + x ( 1 − x ) 3 \dfrac{x^2 + x}{(1 - x)^3} .

∑ n = 1 ∞ n 3 x n = ∑ j = 1 ∞ ∑ n = j ∞ n 2 x n = \sum_{n = 1}^{\infty} n^3 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^2 x^{n} = 1 1 − x ∑ j = 1 ∞ j 2 x j + 2 x ( 1 − x ) 2 ∑ j = 1 ∞ j x j + x 2 + x ( 1 − x ) 3 ∑ j = 1 ∞ x j = ( 1 1 − x ) ( x 2 + x ( 1 − x ) 3 ) + \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^2 x^j + \dfrac{2x}{(1 - x)^2}\sum_{j = 1}^{\infty} j x^j + \dfrac{x^2 + x}{(1 - x)^3}\sum_{j = 1}^{\infty} x^j = (\dfrac{1}{1 - x})(\dfrac{x^2 + x}{(1 - x)^3}) + ( 2 x ( 1 − x ) 2 ) ( x ( 1 − x ) 2 ) + ( x 2 + x ( 1 − x ) 3 ) ( x 1 − x ) = (\dfrac{2x}{(1 - x)^2})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x^2 + x}{(1 - x)^3})(\dfrac{x}{1 - x}) = x 3 + 4 x 2 + x ( 1 − x ) 4 \dfrac{x^3 + 4x^2 + x}{(1 - x)^4} .

g ( x ) = ∑ n = 1 ∞ n 4 x n = ∑ j = 1 ∞ ∑ n = j ∞ n 3 x n = g(x) = \sum_{n = 1}^{\infty} n^4 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^3 x^{n} = 1 1 − x ∑ j = 1 ∞ j 3 x j + 3 x ( 1 − x ) 2 ∑ j = 1 ∞ j 2 x j + 3 ( x 2 + x ) ( 1 − x ) 3 ∑ j = 1 ∞ j x j + x 3 + 4 x 2 + x ( 1 − x ) 4 ∑ j = 1 ∞ x j = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^3 x^j + \dfrac{3x}{(1 - x)^2}\sum_{j = 1}^{\infty} j^2 x^j + \dfrac{3(x^2 + x)}{(1 - x)^3}\sum_{j = 1}^{\infty} j x^j + \dfrac{x^3 + 4x^2 + x}{(1 - x)^4}\sum_{j = 1}^{\infty} x^j =

= ( 1 1 − x ) ( x 3 + 4 x 2 + x ( 1 − x ) 4 ) + = (\dfrac{1}{1 - x})( \dfrac{x^3 + 4x^2 + x}{(1 - x)^4}) + ( 3 x ( 1 − x ) 2 ) ( x 2 + x ( 1 − x ) 3 ) + ( 3 ( x 2 + x ) ( 1 − x ) 3 ) ( x ( 1 − x ) 2 ) + (\dfrac{3x}{(1 - x)^2})(\dfrac{x^2 + x}{(1 - x)^3}) + (\dfrac{3(x^2 + x)}{(1 - x)^3})(\dfrac{x}{(1 - x)^2}) + x 3 + 4 x 2 + x ( 1 − x ) 4 ( x 1 − x ) = x 4 + 11 x 3 + 11 x 2 + x ( 1 − x ) 5 \dfrac{x^3 + 4x^2 + x}{(1 - x)^4}(\dfrac{x}{1 - x}) = \dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5} .

f ( x ) = ∑ n = 1 ∞ n 5 x n = ∑ j = 1 ∞ ∑ n = j ∞ n 4 x n = f(x) = \sum_{n = 1}^{\infty} n^5 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^4 x^{n} =

1 1 − x ∑ j = 1 ∞ j 4 x j + 4 x ( 1 − x ) 2 ∑ j = 1 ∞ j 3 x j + 6 ( x 2 + x ) ( 1 − x ) 3 ∑ j = 1 ∞ j 2 x j + 4 x 3 + 4 x 2 + x ( 1 − x ) 4 ∑ j = 1 ∞ j x j \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^4 x^j + \dfrac{4x}{(1 - x)^2}\sum_{j = 1}^{\infty} j^3 x^j + \dfrac{6(x^2 + x)}{(1 - x)^3}\sum_{j = 1}^{\infty} j^2 x^j + 4\dfrac{x^3 + 4x^2 + x}{(1 - x)^4}\sum_{j = 1}^{\infty} j x^j

+ x 4 + 11 x 3 + 11 x 2 + x ( 1 − x ) 5 ∑ j = 1 ∞ x j = + \dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5}\sum_{j = 1}^{\infty} x^j =

1 1 − x ( x 4 + 11 x 3 + 11 x 2 + x ( 1 − x ) 5 ) \dfrac{1}{1 - x}(\dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5}) + 4 x ( 1 − x ) 2 ( x 3 + 4 x 2 + x ) ( 1 − x ) 4 \dfrac{4x}{(1 - x)^2}(\dfrac{x^3 + 4x^2 + x)}{(1 - x)^4} + 6 ( x 2 + x ) ( 1 − x ) 3 ( x 2 + x ( 1 − x ) 3 ) + 4 ( x 3 + 4 x 2 + x ) ( 1 − x ) 4 ( x ( 1 − x ) 2 ) + ( x 4 + 11 x 3 + 11 x 2 + x ( 1 − x ) 5 ) ( x 1 − x ) \dfrac{6(x^2 + x)}{(1 - x)^3}(\dfrac{x^2 + x}{(1 - x)^3}) + \dfrac{4(x^3 + 4x^2 + x)}{(1 - x)^4}(\dfrac{x}{(1 - x)^2}) + (\dfrac{x^4 + 11x^3 + 11x^2 + x}{(1 - x)^5})(\dfrac{x}{1 - x})

= x 5 + 26 x 4 + 66 x 3 + 26 x 2 + x ( 1 − x ) 6 = \dfrac{x^5 + 26x^4 + 66x^3 + 26x^2 + x}{(1 - x)^6} .

Let u = 1 − x ⟹ d u = − d x ⟹ u = 1 -x \implies du = -dx \implies

∫ − 1 4 0 g ( x ) d x = ∫ 1 5 4 u 4 − 15 u 3 + 50 u 2 − 60 u + 24 u 5 d u = ∫ 1 5 4 1 u − 15 u − 2 + 50 u − 3 − 60 u − 4 + 24 u − 5 d u = \int_{\frac{-1}{4}}^{0} g(x) dx = \int_{1}^{\frac{5}{4}} \dfrac{u^4 - 15u^3 + 50u^2 - 60u + 24}{u^5} du = \int_{1}^{\frac{5}{4}} \dfrac{1}{u} - 15u^{-2} + 50u^{-3} - 60u^{-4} + 24u^{-5} du = ( ln ⁡ ( u ) + 15 u − 25 u 2 + 20 u 3 − 6 u 4 ) ∣ 1 5 4 = (\ln(u) + \dfrac{15}{u} - \dfrac{25}{u^2} + \dfrac{20}{u^3} - \dfrac{6}{u^4})|_{1}^{\frac{5}{4}} = ln ⁡ ( 5 4 ) − 136 625 \ln(\dfrac{5}{4}) - \dfrac{136}{625}

and,

∫ − 1 4 0 f ( x ) d x = ∫ 1 5 4 − u 5 + 31 u 4 − 180 u 3 + 390 u 2 − 360 u + 120 u 6 d u = \int_{\frac{-1}{4}}^{0} f(x) dx = \int_{1}^{\frac{5}{4}} \dfrac{-u^5 + 31u^4 - 180u^3 + 390u^2 - 360u + 120}{u^6} du =

∫ 1 5 4 − 1 u + 31 u − 2 − 180 u − 3 + 390 u − 4 − 360 u − 5 + 120 u − 6 d u = \int_{1}^{\frac{5}{4}} \dfrac{-1}{u} + 31u^{-2} - 180u^{-3} +390u^{-4} - 360u^{-5} + 120u^{-6} du =

( − ln ⁡ ( u ) − 31 u + 90 u 2 − 130 u 3 + 90 u 4 − 24 u 5 ) ∣ 1 5 4 = − ln ⁡ ( 5 4 ) − 14876 3125 + 5 = − ln ⁡ ( 5 4 ) + 749 3125 = 749 3125 − ln ⁡ ( 5 4 ) (-\ln(u) - \dfrac{31}{u} + \dfrac{90}{u^2} - \dfrac{130}{u^3} + \dfrac{90}{u^4} - \dfrac{24}{u^5})|_{1}^{\frac{5}{4}} = -\ln(\dfrac{5}{4}) - \dfrac{14876}{3125} + 5 = -\ln(\dfrac{5}{4}) + \dfrac{749}{3125} = \dfrac{749}{3125} - \ln(\dfrac{5}{4})

⟹ ∫ − 1 4 0 f ( x ) − g ( x ) d x = 1429 3125 − 2 ln ⁡ ( 5 4 ) = 1429 5 5 − 2 ln ⁡ ( 5 2 2 ) = c b b − a ln ⁡ ( b a a ) ⟹ a + b + c = 1436 \implies \int_{\frac{-1}{4}}^{0} f(x) - g(x) dx = \dfrac{1429}{3125} - 2\ln(\dfrac{5}{4}) = \dfrac{1429}{5^5} - 2\ln(\dfrac{5}{2^2}) = \dfrac{c}{b^b} - a\ln(\dfrac{b}{a^a}) \implies a + b + c = \boxed{1436} .

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