Let a_1=b_1=1 and a_n=a_{n-1}+(n-1),b_n=b_{n- | vedclass

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(A) Given $a_1=b_1=1$,$a_n-a_{n-1}=n-1$ and $b_n-b_{n-1}=a_{n-1}$.First,find $a_n$: $a_n = a_1 + \sum_{k=1}^{n-1} k = 1 + \frac{(n-1)n}{2} = \frac{n^2

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$461$

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(A) Given $a_1=b_1=1$,$a_n-a_{n-1}=n-1$ and $b_n-b_{n-1}=a_{n-1}$.First,find $a_n$: $a_n = a_1 + \sum_{k=1}^{n-1} k = 1 + \frac{(n-1)n}{2} = \frac{n^2-n+2}{2}$.For $n=9$,$a_9 = \frac{81-9+2}{2} = 37$.Next,find $b_n$: $b_n = b_1 + \sum_{k=1}^{n-1} a_k = 1 + \sum_{k=1}^{n-1} \frac{k^2-k+2}{2} = 1 + \frac{1}{2} [\frac{(n-1)n(2n-2+1)}{6} - \frac{(n-1)n}{2} + 2(n-1)]$.For $n=10$,$b_{10} = 1 + \sum_{k=1}^{9} a_k = 1 + (1+2+4+7+11+16+22+29+37) = 1 + 129 = 130$.Given $S = \sum_{n=1}^{10} \frac{b_n}{2^n}$ and $T = \sum_{n=1}^8 \frac{n}{2^{n-1}}$.Using the method of differences for $S$,we derive $2S = 2(a_1+b_1) - \frac{b_{10}+2a_9}{2^9} + T$.Thus,$2S - T = 2(1+1) - \frac{130+2(37)}{512} = 4 - \frac{204}{512}$.Multiplying by $2^7 = 128$: $128(4 - \frac{204}{512}) = 512 - \frac{204}{4} = 512 - 51 = 461$.

Let $a_1=b_1=1$ and $a_n=a_{n-1}+(n-1)$,$b_n=b_{n-1}+a_{n-1}$,$\forall n \geq 2$. If $S =\sum \limits_{n=1}^{10} \frac{b_n}{2^n}$ and $T =\sum \limits_{n=1}^8 \frac{n}{2^{n-1}}$,then $2^7(2S - T)$ is equal to $........$.

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