Find the sum of the series up to 10 terms: (3 | vedclass

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(A) The given series is $S = \sum_{n=1}^{10} ((2n+1)^3 - (2n)^3)$.Expanding the terms: $S = (3^3 - 2^3) + (5^3 - 4^3) + \dots + (21^3 - 20^3)$.This ca

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$4960$

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(A) The given series is $S = \sum_{n=1}^{10} ((2n+1)^3 - (2n)^3)$.Expanding the terms: $S = (3^3 - 2^3) + (5^3 - 4^3) + \dots + (21^3 - 20^3)$.This can be written as $S = (3^3 + 5^3 + \dots + 21^3) - (2^3 + 4^3 + \dots + 20^3)$.Adding and subtracting $1^3$ to the first part: $S = (1^3 + 3^3 + \dots + 21^3) - 1^3 - (2^3 + 4^3 + \dots + 20^3)$.We know $\sum_{k=1}^{n} k^3 = [\frac{n(n+1)}{2}]^2$.Sum of odd cubes up to $21^3$: $\sum_{k=1}^{11} (2k-1)^3 = \sum_{k=1}^{22} k^3 - \sum_{k=1}^{11} (2k)^3 = [\frac{22(23)}{2}]^2 - 8[\frac{11(12)}{2}]^2 = 253^2 - 8(66^2) = 64009 - 34848 = 29161$.Sum of even cubes up to $20^3$: $\sum_{k=1}^{10} (2k)^3 = 8 \sum_{k=1}^{10} k^3 = 8 [\frac{10(11)}{2}]^2 = 8(55^2) = 8(3025) = 24200$.$S = 29161 - 24200 = 4961$.Wait,re-evaluating the series: $S = \sum_{n=1}^{10} (8n^3 + 12n^2 + 6n + 1 - 8n^3) = \sum_{n=1}^{10} (12n^2 + 6n + 1)$.$S = 12 \frac{10(11)(21)}{6} + 6 \frac{10(11)}{2} + 10 = 2(10)(11)(21) + 3(110) + 10 = 4620 + 330 + 10 = 4960$.

Find the sum of the series up to $10$ terms: $(3^3 - 2^3) + (5^3 - 4^3) + (7^3 - 6^3) + \dots$

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