**Compute the value of the double summation: ∑i ^ 4 = 1 ∑j=i ^4 (3i + j)**

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## Solution for: Compute the value of the double summation: ∑^{4}_{i = 1}∑^{4}_{j = i} (3i + j)

Evaluating the first sum: ∑^{4}_{j = i}

When j = i in (3i + j)

∑^{4}_{j = i} = (3i +i) + (3i +i) + (3i +i) + (3i +i)

∑^{4}_{j = i} = 4i + 4i + 4i + 4i = 16i

Next, Evaluate the second sum: ∑^{4}_{i = 1} using 16i

when i = 1 in (16i)

∑^{4}_{i = 1} = 16(1) + 16(2) + 16(3) + 16(4)

= 16 + 32 + 48 + 64 = 160

∴ ∑^{4}_{i = 1}∑^{4}_{j = i} (3i + j) = 160