1 ⋅ 4 2 5 + 4 2 ⋅ 7 2 1 1 + 7 2 ⋅ 1 0 2 1 7 + ⋯
Find the value of the closed form of the above series to 2 decimal places.
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We can write the above summation as k = 0 ∑ ∞ ( 3 k + 1 ) 2 ( 3 k + 4 ) 2 ( 3 k + 1 ) + ( 3 k + 4 ) = k = 0 ∑ ∞ ( 3 k + 1 ) 2 ( 3 k + 1 ) 1 + ( 3 k + 1 ) ( 3 k + 4 ) 2 1 = 3 1 k = 0 ∑ ∞ ( 3 k + 1 ) 2 ( 3 k + 4 ) ( 3 k + 4 ) − ( 3 k + 1 ) + ( 3 k + 1 ) ( 3 k + 4 ) 2 ( 3 k + 4 ) − ( 3 k + 1 ) = 3 1 k = 0 ∑ ∞ ( 3 k + 1 ) 2 1 − ( 3 k + 4 ) 2 1 = 3 1 ( 1 − ( 3 ( ∞ ) + 4 ) 2 1 ) = 3 1
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S = 1 ⋅ 4 2 5 + 4 2 ⋅ 7 2 1 1 + 7 2 ⋅ 1 0 2 1 7 + ⋯ = n = 0 ∑ ∞ ( 3 n + 1 ) 2 ( 3 n + 4 ) 2 ( 3 n + 1 ) + ( 3 n + 4 ) = n = 0 ∑ ∞ ( 3 n + 4 ) − ( 3 n + 1 ) ( 3 n + 4 ) − ( 3 n + 1 ) ⋅ ( 3 n + 1 ) 2 ( 3 n + 4 ) 2 ( 3 n + 4 ) + ( 3 n + 1 ) = n = 0 ∑ ∞ 3 1 ⋅ ( 3 n + 1 ) 2 ( 3 n + 4 ) 2 ( 3 n + 4 ) 2 − ( 3 n + 1 ) 2 = 3 1 n = 0 ∑ ∞ ( ( 3 n + 1 ) 2 1 − ( 3 n + 4 ) 2 1 ) = 3 1 ( n = 0 ∑ ∞ ( 3 n + 1 ) 2 1 − n = 1 ∑ ∞ ( 3 n + 1 ) 2 1 ) = 3 1 ⋅ 1 = 3 1 ≈ 0 . 3 3