Let a 1 = b 1 = 1 and a n = a n - 1 + ( n - 1 ) , b n = b n - 1 + a n - 1 , n 2 . If S = n = 1 10 b n 2 n and T = n = 1 8 n 2 n - 1 , then 2 7 ( 2 S - T ) is equal to _________ . [2023]

Question

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 _________ . [2023]

Smart Achievers · Accepted Answer

(461)

$Given,$ $S = \sum_{n = 1}^{10} \frac{b_{n}}{2^{n}} = \frac{b_{1}}{2} + \frac{b_{2}}{2^{2}} + \dots + \frac{b_{9}}{2^{9}} + \frac{b_{10}}{2^{10}}$

$\Rightarrow \frac{S}{2} = \frac{b_{1}}{2^{2}} + \frac{b_{2}}{2^{3}} + \dots + \frac{b_{9}}{2^{10}} + \frac{b_{10}}{2^{11}}$

$Subtracting, we get$

$\frac{S}{2} = \frac{b_{1}}{2} + (\frac{a_{1}}{2^{2}} + \frac{a_{2}}{2^{3}} + \dots + \frac{a_{9}}{2^{10}}) - \frac{b_{10}}{2^{11}}$

$\Rightarrow S = b_{1} - \frac{b_{10}}{2^{10}} + (\frac{a_{1}}{2} + \frac{a_{2}}{2^{2}} + \dots + \frac{a_{9}}{2^{9}})$

$\Rightarrow \frac{S}{2} = \frac{b_{1}}{2} - \frac{b_{10}}{2^{11}} + (\frac{a_{1}}{2^{2}} + \frac{a_{2}}{2^{3}} + \dots + \frac{a_{9}}{2^{10}})$

$Subtracting, we get:$

$\frac{S}{2} = \frac{b_{1}}{2} - \frac{b_{10}}{2^{11}} + (\frac{a_{1}}{2} - \frac{a_{9}}{2^{10}}) + (\frac{1}{2^{2}} + \frac{2}{2^{3}} + \dots + \frac{8}{2^{9}})$

$\Rightarrow \frac{S}{2} = \frac{a_{1} + b_{1}}{2} - \frac{(b_{10} + 2 a_{9})}{2^{11}} + \frac{T}{4}$

$\Rightarrow 2 S = 2 (a_{1} + b_{1}) - \frac{b_{10} + 2 a_{9}}{2^{9}} + T$

$\Rightarrow 2^{7} (2 S - T) = 2^{8} (a_{1} + b_{1}) - \frac{(b_{10} + 2 a_{9})}{4}$

$Given a_{n} - a_{n - 1} = n - 1$

$∴$ $\begin{matrix} a_{2} & - & a_{1} & = & 1 \\ a_{3} & - & a_{2} & = & 2 \\ ⋮ \\ a_{9} & - & a_{8} & = & 8 \end{matrix}$

_________________

$a_{9} - a_{1} = 1 + 2 + \dots 8 = 36$

$\Rightarrow a_{9} = 37$

$Also, b_{n} - b_{n - 1} = a_{n - 1}$ $∴$ $b_{10} - b_{1} = a_{1} + a_{2} + \dots + a_{9}$

$= 1 + 2 + 4 + 7 + 11 + 16 + 22 + 29 + 37$

$\Rightarrow b_{10} = 130$

$∴$ $2^{7} (2 S - T) = 2^{8} (1 + 1) - \frac{(130 + 2 \times 37)}{4}$ = 461

Solution

Q.
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 _________ . [2023]

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Solution

Q. 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=18n2n-1, then 27(2S-T) is equal to _________ . [2023]

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Q.
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 _________ . [2023]