Dual summation

Calculus Level 4

{ N = ∑ j = 1 1728 j ( j − 1727 ) M = ∑ k = 1 ∞ 6 k + 7 7 k + 1 k ( k + 1 ) \begin{cases} \displaystyle N = \sum_{j=1}^{1728} j(j-1727) \\ \displaystyle M = \sum_{k=1}^{\infty}\frac{6k+7}{7^{k+1}k(k+1)}\end{cases}

For N N and M M as defined above, find the remainder when M N + 1 M \dfrac{MN+1}{M} is divided by 1000.

The problem is original.


The answer is 759.

Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

M = ∑ k = 1 ∞ 6 k + 7 7 k + 1 k ( k + 1 ) = ∑ k = 1 ∞ 6 ( k + 1 ) + 1 7 k + 1 k ( k + 1 ) = ∑ k = 1 ∞ 1 7 k + 1 ( 6 k + 1 k ( k + 1 ) ) = ∑ k = 1 ∞ 1 7 k + 1 ( 6 k + 1 k − 1 k + 1 ) = ∑ k = 1 ∞ 1 7 k + 1 ( 7 k − 1 k + 1 ) = ∑ k = 1 ∞ ( 1 7 k k − 1 7 k + 1 ( k + 1 ) ) = 1 7 \begin{aligned} M & = \sum_{k=1}^\infty \frac {6k+7}{7^{k+1}k(k+1)} \\ & = \sum_{k=1}^\infty \frac {6(k+1)+1}{7^{k+1}k(k+1)} \\ & = \sum_{k=1}^\infty \frac 1{7^{k+1}}\left(\frac 6k + \frac 1{k(k+1)} \right) \\ & = \sum_{k=1}^\infty \frac 1{7^{k+1}}\left(\frac 6k + \frac 1k - \frac 1{k+1} \right) \\ & = \sum_{k=1}^\infty \frac 1{7^{k+1}}\left(\frac 7k - \frac 1{k+1} \right) \\ & = \sum_{k=1}^\infty \left(\frac 1{7^kk} - \frac 1{7^{k+1}(k+1)} \right) \\ & = \frac 17 \end{aligned}

N = ∑ j = 1 n j ( j − ( n − 1 ) ) Let n = 1728 = ∑ j = 1 n j 2 − ( n − 1 ) ∑ j = 1 n j = n ( n + 1 ) ( 2 n + 1 ) 6 − ( n − 1 ) ⋅ n ( n + 1 ) 2 = n ( n + 1 ) 2 ( 2 n + 1 3 − ( n − 1 ) ) = − n ( n + 1 ) ( n − 4 ) 6 = − 1728 ( 1729 ) ( 1724 ) 6 = − 288 ( 1729 ) ( 1724 ) \begin{aligned} N & = \sum_{j=1}^{\color{#3D99F6}n} j(j-(n-1)) & \small \color{#3D99F6} \text{Let }n=1728 \\ & = \sum_{j=1}^n j^2 -(n-1) \sum_{j=1}^n j \\ & = \frac {n(n+1)(2n+1)}6 - (n-1)\cdot \frac {n(n+1)}2 \\ & = \frac {n(n+1)}2\left(\frac {2n+1}3-(n-1) \right) \\ & = - \frac {n(n+1)(n-4)}6 \\ & = - \frac {1728(1729)(1724)}6 \\ & = - 288(1729)(1724) \end{aligned}

Therefore,

M N + 1 M ≡ N + 1 M (mod 1000) ≡ − 288 ( 1729 ) ( 1724 ) + 7 (mod 1000) ≡ − 288 ( 729 ) ( 724 ) + 7 (mod 1000) ≡ − 288 ( 271 ) ( 276 ) + 7 (mod 1000) ≡ − 288 ( 250 + 21 ) ( 250 + 26 ) + 7 (mod 1000) ≡ − 18 ( 1000 + 84 ) ( 1000 + 69 ) + 7 (mod 1000) ≡ − 18 ( 84 ) ( 104 ) + 7 (mod 1000) ≡ − 18 ( 100 − 16 ) ( 100 + 4 ) + 7 (mod 1000) ≡ − 18 ( − 1264 ) + 7 (mod 1000) ≡ 18 ( 264 ) + 7 (mod 1000) ≡ 4752 + 7 (mod 1000) ≡ 759 (mod 1000) \begin{aligned} \frac {MN+1}M & \equiv N + \frac 1M \text{ (mod 1000)} \\ & \equiv - 288(1729)(1724) + 7 \text{ (mod 1000)} \\ & \equiv - 288(729)(724) + 7 \text{ (mod 1000)} \\ & \equiv - 288(271)(276) + 7 \text{ (mod 1000)} \\ & \equiv - 288(250+21)(250+26) + 7 \text{ (mod 1000)} \\ & \equiv - 18(1000+84)(1000+69) + 7 \text{ (mod 1000)} \\ & \equiv - 18(84)(104) + 7 \text{ (mod 1000)} \\ & \equiv - 18(100-16)(100+4) + 7 \text{ (mod 1000)} \\ & \equiv - 18(-1264) + 7 \text{ (mod 1000)} \\ & \equiv 18(264) + 7 \text{ (mod 1000)} \\ & \equiv 4752 + 7 \text{ (mod 1000)} \\ & \equiv \boxed{759} \text{ (mod 1000)} \end{aligned}

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