JEE Main 2023 — Mathematics Algebra

Given:

$${a}_{n}={a}_{n-1}+(n-1)$$an​=an−1​+(n−1)

$${b}_{n}={b}_{n-1}+{a}_{n-1}$$bn​=bn−1​+an−1​

And,

$$S=\frac{{b}_{1}}{2}+\frac{{b}_{2}}{{2}^{2}}+\ldots \ldots ..+\frac{{b}_{9}}{{2}^{9}}+\frac{{b}_{10}}{{2}^{10}}...(1)$$S=2b1​​+22b2​​+……..+29b9​​+210b10​​...(1)

$$\Rightarrow \frac{S}{2}=\frac{{b}_{1}}{{2}^{2}}+\frac{{b}_{2}}{{2}^{3}}+\ldots \ldots ..+\frac{{b}_{9}}{{2}^{10}}+\frac{{b}_{10}}{{2}^{11}}...(2)$$⇒2S​=22b1​​+23b2​​+……..+210b9​​+211b10​​...(2)

Subtracting $(1)$(1) and $(2)$(2), we get

$$\frac{S}{2}=\frac{b}{2}+(\frac{{b}_{2}-{b}_{1}}{{2}^{2}})+(\frac{{b}_{3}-{b}_{2}}{{2}^{3}})+....+(\frac{{b}_{10}-{b}_{9}}{{2}^{10}})-(\frac{{b}_{10}}{{2}^{11}})$$2S​=2b​+(22b2​−b1​​)+(23b3​−b2​​)+....+(210b10​−b9​​)−(211b10​​)

$$\Rightarrow \frac{S}{2}=\frac{{b}_{1}}{2}+(\frac{{a}_{1}}{{2}^{2}}+\frac{{a}_{2}}{{2}^{3}}\ldots \ldots .+\frac{{a}_{9}}{{2}^{10}})-\frac{{b}_{10}}{{2}^{11}}$$⇒2S​=2b1​​+(22a1​​+23a2​​…….+210a9​​)−211b10​​

$$\Rightarrow S={b}_{1}-\frac{{b}_{10}}{{2}^{10}}+(\frac{{a}_{1}}{2}+\frac{{a}_{2}}{{2}^{2}}\ldots \ldots ..+\frac{{a}_{9}}{{2}^{9}})....(3)$$⇒S=b1​−210b10​​+(2a1​​+22a2​​……..+29a9​​)....(3)

$$\Rightarrow \frac{S}{2}=\frac{{b}_{1}}{2}-\frac{{b}_{10}}{{2}^{11}}+(\frac{{a}_{1}}{{2}^{2}}+\frac{{a}_{2}}{{2}^{3}}\ldots \ldots .+\frac{{a}_{9}}{{2}^{10}})...(4)$$⇒2S​=2b1​​−211b10​​+(22a1​​+23a2​​…….+210a9​​)...(4)

$$\Rightarrow \frac{S}{2}=\frac{{b}_{1}}{2}-\frac{{b}_{10}}{{2}^{11}}+(\frac{{a}_{1}}{2}-\frac{{a}_{9}}{{2}^{10}})+(\frac{1}{{2}^{2}}+\frac{2}{{2}^{3}}+\ldots +\frac{8}{{2}^{9}})$$⇒2S​=2b1​​−211b10​​+(2a1​​−210a9​​)+(221​+232​+…+298​)

$$\Rightarrow \frac{S}{2}=\frac{{a}_{1}+{b}_{1}}{2}-\frac{({b}_{10}+2{a}_{9})}{{2}^{11}}+\frac{T}{4}$$⇒2S​=2a1​+b1​​−211(b10​+2a9​)​+4T​

$$\Rightarrow 2S=2({a}_{1}+{b}_{1})-\frac{({b}_{10}+2{a}_{9})}{{2}^{9}}+T$$⇒2S=2(a1​+b1​)−29(b10​+2a9​)​+T

$$\Rightarrow (2S-T)=2({a}_{1}+{b}_{1})-\frac{({b}_{10}+2{a}_{9})}{{2}^{9}}$$⇒(2S−T)=2(a1​+b1​)−29(b10​+2a9​)​

$$\Rightarrow {2}^{7}(2S-T)={2}^{8}({a}_{1}+{b}_{1})-\frac{({b}_{10}+2{a}_{9})}{4}$$⇒27(2S−T)=28(a1​+b1​)−4(b10​+2a9​)​

Given,

$${a}_{n}-{a}_{n-1}=n-1$$an​−an−1​=n−1

$${a}_{2}-{a}_{1}=1$$a2​−a1​=1

$${a}_{3}-{a}_{2}=2$$a3​−a2​=2

$$\vdots \vdots \vdots \vdots \vdots \vdots \vdots \vdots \vdots$$⋮⋮⋮⋮⋮⋮⋮⋮⋮

$${a}_{9}-{a}_{8}=8$$a9​−a8​=8

Adding all equations, we get

$${a}_{9}-{a}_{1}=1+2+\ldots +8=36$$a9​−a1​=1+2+…+8=36

$$\Rightarrow {a}_{9}=37(\because {a}_{1}=1)$$⇒a9​=37(∵a1​=1)

So,

$${2}^{7}(2S-T)={2}^{9}-\frac{({b}_{10}+2\cdot 37)}{4}$$27(2S−T)=29−4(b10​+2⋅37)​

Also,

$${b}_{n}-{b}_{n-1}={a}_{n-1}$$bn​−bn−1​=an−1​

$$\therefore {b}_{2}-{b}_{1}={a}_{1}$$∴b2​−b1​=a1​

$${b}_{3}-{b}_{2}={a}_{2}$$b3​−b2​=a2​

$${b}_{4}-{b}_{3}={a}_{3}$$b4​−b3​=a3​

$$\vdots \vdots \vdots \vdots \vdots \vdots \vdots \vdots \vdots$$⋮⋮⋮⋮⋮⋮⋮⋮⋮

$${b}_{10}-{b}_{9}={a}_{9}$$b10​−b9​=a9​

Adding all equations, we get

$${b}_{10}-{b}_{1}={a}_{1}+{a}_{2}+...+{a}_{9}$$b10​−b1​=a1​+a2​+...+a9​

$$\Rightarrow {b}_{10}-{b}_{1}=1+2+4+7+11+16+22+29+37$$⇒b10​−b1​=1+2+4+7+11+16+22+29+37

$$\Rightarrow {b}_{10}=130$$⇒b10​=130

So,

$${2}^{7}(2S-T)={2}^{9}-\frac{({b}_{10}+2\cdot 37)}{4}={2}^{9}-\frac{(130+2\cdot 37)}{4}=461$$27(2S−T)=29−4(b10​+2⋅37)​=29−4(130+2⋅37)​=461