Let a1=b1=1 and an=an−1+(n−1),bn=bn−1+an−1,∀n≥2. If S=∑n=110 bn2n and T=∑n=18 n2n−1then 27(2S−T) is equal to ________.

S=∑n=110 bn2n=b12+b222+…….+b10210S2=b122+….+b7210+b10211 on subtracting S2=b12+b2−b122+b3−b223+…….+b10−b9210−b10211S2=12+a122+a223+…+a9210−b10211S4=14+a123+a224+……+a8210+a9211−b10212 on subtracting S4=14+a2−a123+a3−a224+…..+a9−a8210−b10211−a9211+b10212S=1+a2−a12+a3−a222+….+a9−a828−a929−b10210S=1+12+222+323+….+828−a929−b10210T=1+22+322+….+8272S=2+1+221+322+….+827−a928−b10292S−T=2−a928−b102927(2S−T)=28−a92−b104………..(1)an−an−1=n−1ak=ak2+bk+ca+b+c=14a+2b+c=29a+3b+c=4a=12,b=−12,c=1a9=37bk=ak3+bk2+ck+da+b+c+d=18a+4b+2c+d=227a+9b+3c+d=464a+16b+4c+d=8bk=k36−k22+43kb10=130Using a9& b10 in equation ….(1)We get S=461

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

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answer is 461.

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Detailed Solution

$\begin{array}{l} S = \sum_{n = 1}^{10} \frac{b_{n}}{2^{n}} = \frac{b_{1}}{2} + \frac{b_{2}}{2^{2}} + \dots \dots . + \frac{b_{10}}{2^{10}} \\ \frac{S}{2} = \frac{b_{1}}{2^{2}} + \dots . + \frac{b_{7}}{2^{10}} + \frac{b_{10}}{2^{11}} on subtracting \\ \frac{S}{2} = \frac{b_{1}}{2} + \frac{b_{2} - b_{1}}{2^{2}} + \frac{b_{3} - b_{2}}{2^{3}} + \dots \dots . + \frac{b_{10} - b_{9}}{2^{10}} - \frac{b_{10}}{2^{11}} \\ \frac{S}{2} = \frac{1}{2} + \frac{a_{1}}{2^{2}} + \frac{a_{2}}{2^{3}} + \dots + \frac{a_{9}}{2^{10}} - \frac{b_{10}}{2^{11}} \\ \frac{S}{4} = \frac{1}{4} + \frac{a_{1}}{2^{3}} + \frac{a_{2}}{2^{4}} + \dots \dots + \frac{a_{8}}{2^{10}} + \frac{a_{9}}{2^{11}} - \frac{b_{10}}{2^{12}} on subtracting \\ \frac{S}{4} = \frac{1}{4} + \frac{a_{2} - a_{1}}{2^{3}} + \frac{a_{3} - a_{2}}{2^{4}} + \dots . . + \frac{a_{9} - a_{8}}{2^{10}} - \frac{b_{10}}{2^{11}} - \frac{a_{9}}{2^{11}} + \frac{b_{10}}{2^{12}} \\ S = 1 + \frac{a_{2} - a_{1}}{2} + \frac{a_{3} - a_{2}}{2^{2}} + \dots . + \frac{a_{9} - a_{8}}{2^{8}} - \frac{a_{9}}{2^{9}} - \frac{b_{10}}{2^{10}} \\ S = 1 + \frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + \dots . + \frac{8}{2^{8}} - \frac{a_{9}}{2^{9}} - \frac{b_{10}}{2^{10}} \end{array}$
$\begin{array}{l} T = 1 + \frac{2}{2} + \frac{3}{2^{2}} + \dots . + \frac{8}{2^{7}} \\ 2 S = 2 + (1 + \frac{2}{2^{1}} + \frac{3}{2^{2}} + \dots . + \frac{8}{2^{7}}) - \frac{a_{9}}{2^{8}} - \frac{b_{10}}{2^{9}} \\ 2 S - T = 2 - \frac{a_{9}}{2^{8}} - \frac{b_{10}}{2^{9}} \\ 2^{7} (2 S - T) = 2^{8} - \frac{a_{9}}{2} - \frac{b_{10}}{4} \dots \dots \dots . . (1) \\ a_{n} - a_{n - 1} = n - 1 \\ a_{k} = {ak}^{2} + bk + c \\ a + b + c = 1 \\ 4 a + 2 b + c = 2 \\ 9 a + 3 b + c = 4 \\ a = \frac{1}{2}, b = - \frac{1}{2}, c = 1 \\ a_{9} = 37 \end{array}$
$\begin{array}{l} b_{k} = {ak}^{3} + {bk}^{2} + ck + d \\ a + b + c + d = 1 \\ 8 a + 4 b + 2 c + d = 2 \\ 27 a + 9 b + 3 c + d = 4 \\ 64 a + 16 b + 4 c + d = 8 \\ b_{k} = \frac{k^{3}}{6} - \frac{k^{2}}{2} + \frac{4}{3} k \\ b_{10} = 130 \end{array}$
Using a₉& b₁₀ in equation ….(1)
We get S=461