Multiple Summations

Algebra Level 3

i = 0 2 j = 0 3 ( 2 i + 3 j ) = ? \large\displaystyle \sum _{ i=0 }^{ 2 } \sum _{ j=0 }^{ 3 } (2i+3j)=?


The answer is 78.

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2 solutions

S = i = 0 2 j = 0 3 ( 2 i + 3 j ) Note that i is independent of j and is treated as a constant. = i = 0 2 ( 2 i ( 3 + 1 ) + 3 ( 3 ) ( 3 + 1 ) 2 ) = i = 0 2 ( 8 i + 18 ) = 8 ( 2 ) ( 2 + 1 ) 2 + 18 ( 2 + 1 ) = 24 + 54 = 78 \begin{aligned} S & = \sum_{i=0}^2 \sum_{j=0}^{\color{#D61F06}3} (2{\color{#3D99F6}i}+3j) & \small \color{#3D99F6} \text{Note that } i \text{ is independent of } j \text{ and is treated as a constant.} \\ & = \sum_{i=0}^{\color{#3D99F6}2} \left(2 {\color{#3D99F6}i} ({\color{#D61F06}3}+1)+\frac{3({\color{#D61F06}3})({\color{#D61F06}3}+1)}2 \right) \\ & = \sum_{i=0}^{\color{#3D99F6}2} \left(8 {\color{#3D99F6}i} + 18 \right) \\ & = \frac{8 ({\color{#3D99F6}2})({\color{#3D99F6}2}+1)}2 + 18({\color{#3D99F6}2}+1) \\ & = 24 + 54 = \boxed{78} \end{aligned}

Precisely, \text{Precisely,}

S = i = 0 2 j = 0 3 ( 2 i + 3 j ) \large \mathfrak{S} = \sum_{i=0}^{2} \sum_{j=0}^{3} (2i+3j)

S = i = 0 2 ( 2 i + 2 i + 3 + 2 i + 6 + 2 i + 9 ) \large \mathfrak{S} = \sum_{i=0}^{2} (2i + 2i + 3 + 2i + 6 + 2i + 9)

S = i = 0 2 ( 8 i + 18 ) \large \mathfrak{S} = \sum_{i=0}^{2} (8i+18)

S = 8.3 + 18.3 = 78 \large \mathfrak{S} = 8.3 + 18.3 = \boxed{78}

Typo: 24 + 54 = 78 24 + 54 = 78

Swapnil Das - 5 years, 2 months ago

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Thanks, I've edited that. Check out I've mentioned you in a note

Aditya Narayan Sharma - 5 years, 2 months ago

Swapnil in which class you r studing..............

Abhisek Mohanty - 5 years, 2 months ago

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