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Question Number 7163 by Tawakalitu. last updated on 14/Aug/16

Commented by sou1618 last updated on 14/Aug/16
$Σ_(a=1) ^4 (Σ_(b=0) ^4 a^b )=Σ_(a=1) ^4 (a^0 +a^1 +a^2 +a^3 +a^4 ) Σ_(a=1) ^4 (1+a+a^2 +a^3 +a^4 ) Σk=(1/2)n(n+1) Σk^2 =(1/6)n(n+1)(2n+1) Σk^3 =(1/4){n(n+1)}^2 Σk^4 =(1/(30))n(n+1)(2n+1)(3n^2 +3n−1) =4+10+(1/6)4(4+1)(2×4+1)+(1/4){4(4+1)}^2 +(1/(30))4(4+1)(8+1)(3×16+3×4−1) =14+30+100+6×59 =498$
$\underset{a = 1}{\sum^{4}} (\underset{b = 0}{\sum^{4}} a^{b}) = \underset{a = 1}{\sum^{4}} (a^{0} + a^{1} + a^{2} + a^{3} + a^{4})$ $\underset{a = 1}{\sum^{4}} (1 + a + a^{2} + a^{3} + a^{4})$ $Σ k = \frac{1}{2} n (n + 1)$ $Σ k^{2} = \frac{1}{6} n (n + 1) (2 n + 1)$ $Σ k^{3} = \frac{1}{4} {n (n + 1)}^{2}$ $Σ k^{4} = \frac{1}{30} n (n + 1) (2 n + 1) (3 n^{2} + 3 n - 1)$ $= 4 + 10 + \frac{1}{6} 4 (4 + 1) (2 \times 4 + 1) + \frac{1}{4} {4 (4 + 1)}^{2} + \frac{1}{30} 4 (4 + 1) (8 + 1) (3 \times 16 + 3 \times 4 - 1)$ $= 14 + 30 + 100 + 6 \times 59$ $= 498$
Commented by Tawakalitu. last updated on 14/Aug/16

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