Question

Evaluate : $\mathtt {\dfrac{2^n + 2^{n-1}}{2^{n+1}-2^n}}$​

Answer 1

✠ Yᴏᴜʀ Qᴜᴇsᴛɪᴏɴ ☟

$\\ \qquad ☯ \: \: \bf \underline{ Evaluate }: - \: \mathtt {\dfrac{2^n + 2^{n-1}}{2^{n+1}-2^n}}$

✠ Rᴇǫᴜɪʀᴇᴅ Aɴsᴡᴇʀ ☟

$\\ \bigstar \: \: \underline {\boxed{\bf { \color{purple}\: \underline{ Question} \: : - \: \mathtt {\dfrac{2^n + 2^{n-1}}{2^{n+1}-2^n}}}}} \: \: \bigstar$

$: \implies \: \mathtt {\dfrac{ \cancel{2^n} (1 + 2^{-1})}{ \cancel{2^{n}} (2- 1)}}$

$༒ \: \: \frak{ \: \underline {There \: 2 \: to \: the \: power \: of \: n \: is \: cancelled.}}$

$: \implies \: \tt \frac{1 + 2 {}^{ - 1} }{1}$

$༒ \: \: \frak{ \: \underline {Calculate \: 2 \: t o \: the \: power \: of \: -1 \: and \: get \: \dfrac{1}{2}.}}$

$: \implies \: \tt\frac{1 + \frac{1}{2} }{1}$

$༒ \: \: \frak{ \: \underline {Add \: 1 \: and \: \frac{1}{2} \: to \: get \: \frac{3}{2}. }}$

$: \implies \: \tt \frac{ \frac{3}{2} }{1}$

$༒ \: \: \frak{ \: \underline {Anything \: divided \: by \: one \: gives \: itself.}}$

$\bigstar \: \: \underline{ \boxed{ \color{purple}{ \bf \: Answer : ➠ \: \: \tt \frac{3}{2} \approx \: \bf \: 1.5}}} \: \: \bigstar$

✯ Hope it helps u ✯

Answer 2

Step-by-step explanation:

Given:-

[2^n +2^(n-1)]/[2^(n+1)-2^n]

To find:-

Evaluate [2^n +2^(n-1)]/[2^(n+1)-2^n]

Solution:-

Given that

[2^n +2^(n-1)]/[2^(n+1)-2^n]

we know that a^(m-n)= a^m/a^n

and a^(m+n)=a^m×a^n

=>[2^n+(2^n/2^1)]/[(2^n×2^1)-2^n]

=>2^n[1+(1/2)] / [2^n(2-1)]

On cancelling 2^n in numerator and denominator

=>[1+(1/2)]/(2-1)

=>[(2+1)/2]/1

=>3/2

Answer:-

The value of the given expression is 3/2

Used formulae:-

• a^(m-n)= a^m/a^n

• a^(m+n)=a^m×a^n