Question

The interval in which x
must lie so that the numerically greatest term in the expansion of (1-x)²¹ has the greatest coefficient is (a/b,6/5) Find the value of (a+b).

Please provide enough explanation.
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

The interval in which x must lie so that the numerically greatest term in the expansion of (1-x)²¹ has the greatest coefficient is (a/b,6/5).

Now, the given expansion is

$\rm \: {(1 - x)}^{21} \: \: provided \: that \: x > 0$

We know, in the expansion of

$\rm \: {(1 - x)}^{n} \: where \: n \: is \: odd, \: the \: terms \: having \: maximum \: numerical \: value \: is \:$

$\boxed{\sf{  \:\rm \:T_{\dfrac{n + 1}{2} } \: \: and \: \:T_{\dfrac{n + 3}{2} } }}$

So, here n = 21, which is odd.

So, terms having maximum numeric value are

$\rm \: T_{\dfrac{21 + 1}{2} } \: \: and \: \: T_{\dfrac{21 + 3}{2} }$

$\rm \: T_{\dfrac{22}{2} } \: \: and \: \: T_{\dfrac{24}{2} }$

$\rm\implies \:T_{11} \: \: and \: T_{12}$

So, we have now

$\rm \: |T_{11}| > |T_{10}| \: \: and \: \: |T_{12}| > |T_{13}|$

$\rm \: |T_{10 + 1}| > |T_{9 + 1}| \: \: and \: \: |T_{11 + 1}| > |T_{12 + 1}|$

$\rm \: |^{21}C_{10} {( - x)}^{10} | > |^{21}C_{9} {( - x)}^{9} | \: \: and \: \: |^{21}C_{11} {( - x)}^{11} | > |^{21}C_{12} {( - x)}^{12} |$

$\rm \: |^{21}C_{10} {x}^{10} | > |^{21}C_{9} {x}^{9} | \: \: and \: \: |^{21}C_{11} { x}^{11} | > |^{21}C_{12} {x}^{12} |$

$\rm \: \dfrac{^{21}C_{10}}{^{21}C_{9}} x > 1 \: \: and \: \: \dfrac{^{21}C_{12}}{^{21}C_{11}} x < 1$

We know

$\boxed{\sf{  \:\rm \: \dfrac{^{n}C_{r}}{^{n}C_{r - 1}} = \frac{n - r + 1}{r} \: }}$

So, using this identity, we get

$\rm \: \dfrac{21 - 9}{10} \: x > 1 \: \: and \: \: \dfrac{21 - 11}{12} \: x < 1$

$\rm \: \dfrac{12x}{10} \: > 1 \: \: and \: \: \dfrac{10x}{12} \: < 1$

$\rm \: \dfrac{6x}{5} \: > 1 \: \: and \: \: \dfrac{5x}{6} \: < 1$

$\rm\implies \:x > \dfrac{5}{6} \: \: and \: \: x < \dfrac{6}{5}$

$\rm\implies \:x \: \in \: \bigg(\dfrac{5}{6}, \: \dfrac{6}{5} \bigg)$

As it is given that,

$\rm \: \:x \: \in \: \bigg(\dfrac{a}{b}, \: \dfrac{6}{5} \bigg)$

So, on comparing, we get

$\rm \: a \: = \: 5 \: \: and \: \: b \: = \: 6$

$\rm\implies \:a + b = 5 + 6 = 11$

$\rule{190pt}{2pt}$

Formula used :-

$\sf \: In \: the \: expansion \: of \: {(x + y)}^{n}, \: the \: general \: term \: is \:$

$\boxed{\sf{  \:\rm \: T_{r + 1} \: = \: ^{n}C_{r} \: {x}^{n - r} \: {y}^{r} \: \: }}$