Question

question no. 13 plz solve.​

Answer 1

Solution:-

Given

$\rm \to \: S_{n} = 1 + q + {q}^{2} + ......... + {q}^{n}$

$\rm \to \: T_{n} = 1 + \bigg( \dfrac{q + 1}{2} \bigg) + \bigg( \dfrac{q + 1}{2} \bigg)^{2} + ........ + \bigg( \dfrac{q + 1}{2} \bigg)^{n}$

$\rm \to \: q \: is \: real \: number \: and \: q \not = 0$

$\rm \to \: ^{101} C_{1} + ^{101} C_{2}.S_{1} + ...... + ^{101} C_{101}.S _{100} = \alpha T_{100}$

To find

$\to \: \rm \: to \: find \: the \: \: value \: of \: alpha( \alpha )$

Now take

$\rm \to \: S_{n} = 1 + q + {q}^{2} + ......... + {q}^{n}$

So total number of term is ( n + 1 )

We can write as

$\rm \to \: \dfrac{q^{n + 1} - 1}{q - 1}$

Now take

$\rm \to \: T_{n} = 1 + \bigg( \dfrac{q + 1}{2} \bigg) + \bigg( \dfrac{q + 1}{2} \bigg)^{2} + ........ + \bigg( \dfrac{q + 1}{2} \bigg)^{n}$

Again total number of term is ( n + 1 )

$\rm \to \: \dfrac{ \bigg( \dfrac{q + 1}{2} \bigg)^{n + 1} - 1}{ \bigg( \dfrac{q + 1}{2} \bigg) - 1 }$

Now

$\rm \to \: ^{101} C_{1} + ^{101} C_{2}.S_{1} + ...... + ^{101} C_{101}.S _{100} = \alpha T_{100}$

We can write

$\rm \to \alpha T_{101} = \sum \limits_{r= 1}^{100} \: ^{101} C_rS_{r - 1}$

When we take r - 1

$\rm \to\sum \limits_{r - 1}^{101} \: ^{101} C_r \bigg( \dfrac{ {q}^{r - 1} }{q - 1} \bigg)$

$\rm \to \dfrac{1}{q - 1} \bigg(\sum \limits_{r - 1}^{101} \: ^{101} C_{r} {q}^{r} - \sum \limits_{r - 1}^{101} \: ^{101} C_{r} \bigg)$

$\rm \to \: \dfrac{1}{q - 1} \bigg \{(1 + q) {}^{101 } -1 - ({2}^{100} - 1) \bigg\}$

$\rm \to \: \dfrac{1}{q - 1} \bigg \{(1 + q) {}^{101} - {2}^{101} \bigg \}$

Now we get

$\rm \to \: \dfrac{ \bigg( \dfrac{q + 1}{2} \bigg)^{n + 1} - 1}{ \bigg( \dfrac{q + 1}{2} \bigg) - 1 }$

Now put the value n = 100

$\rm \to \: \alpha \bigg( \dfrac{ \bigg( \dfrac{q + 1}{2} \bigg) {}^{101} - 1}{ \dfrac{q + 1}{2} - 1 } \bigg) = \dfrac{(1 + q) ^{101} - {2}^{101} }{q - 1}$

By taking lcm we get

$\rm \to \: \dfrac{2}{ {2}^{101} } \alpha \bigg( \dfrac{(q + 1) {}^{101} - {2}^{101} }{(q + 1) - 2} \bigg) = \dfrac{(1 + q) {}^{101} - {2}^{101} }{q - 1}$

$\rm \to \: \dfrac{2}{ {2}^{101} } \alpha \bigg( \dfrac{(q + 1) {}^{101} - {2 }^{101} }{(q - 1) } \bigg) = \dfrac{(1 + q) {}^{101} - {2}^{101} }{q - 1}$

$\rm \to \: \dfrac{2}{ {2}^{101} } \alpha \cancel{\bigg( \dfrac{(q + 1) {}^{101} - {2}^{101} }{(q + 1) - 2} \bigg) }= \cancel{ \dfrac{(1 + q) {}^{101} - {2}^{101} }{q - 1} }$

$\rm \to \: \alpha \bigg( \dfrac{1}{ {2}^{100} } \bigg) = 1$

$\rm \to \: \alpha = {2}^{100}$

Answer:- Option A is correct

Answer 2

Option A is correct............!