If 1+(1-2^{2} cdot 1)+(1-4^{2} cdot 3)+(1-6^{ | vedclass

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(B) The given expression is $S = 1 + \sum_{n=1}^{10} (1 - (2n)^2(2n-1))$.There are $10$ terms in the summation plus the initial $1$,making a total of

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$(11, 103)$

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(B) The given expression is $S = 1 + \sum_{n=1}^{10} (1 - (2n)^2(2n-1))$.There are $10$ terms in the summation plus the initial $1$,making a total of $11$ terms.$S = 11 - \sum_{n=1}^{10} (4n^2)(2n-1) = 11 - 4 \sum_{n=1}^{10} (2n^3 - n^2)$.Using the summation formulas $\sum n^3 = [\frac{n(n+1)}{2}]^2$ and $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ for $n=10$:$\sum_{n=1}^{10} n^3 = (55)^2 = 3025$.$\sum_{n=1}^{10} n^2 = \frac{10 	imes 11 	imes 21}{6} = 385$.Substituting these values:$S = 11 - 4 [2(3025) - 385] = 11 - 4 [6050 - 385] = 11 - 4(5665) = 11 - 22660$.We need the form $\alpha - 220 \beta$. Since $22660 = 220 	imes 103$,we have $S = 11 - 220(103)$.Thus,$(\alpha, \beta) = (11, 103)$.

If $1+(1-2^{2} \cdot 1)+(1-4^{2} \cdot 3)+(1-6^{2} \cdot 5)+\ldots+(1-20^{2} \cdot 19) = \alpha - 220 \beta$,then an ordered pair $(\alpha, \beta)$ is equal to

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