$$ 2 x+1)^{4}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4} $$ रून, $$ a_{0}-2 a_{1}+4 a_{2}-8 a_{3}+16 a $$

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To find the coefficients $$ a_0, a_1, a_2, a_3, a_4 $$ in the expansion of $$ (2x+1)^4 $$, we can use the binomial theorem.

The binomial theorem states that for any positive integer $$ n $$, the expansion of $$ (a+b)^n $$ is given by:
$$ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k $$

In this case, we have $$ a = 2x $$ and $$ b = 1 $$, and we want to find the expansion of $$ (2x+1)^4 $$.

Using the binomial theorem, we can expand $$ (2x+1)^4 $$ as:
$$ (2x+1)^4 = \binom{4}{0} (2x)^4 + \binom{4}{1} (2x)^3 (1) + \binom{4}{2} (2x)^2 (1)^2 + \binom{4}{3} (2x) (1)^3 + \binom{4}{4} (1)^4 $$

Simplifying each term, we get:
$$ (2x+1)^4 = 16x^4 + 32x^3 + 24x^2 + 8x + 1 $$

Therefore, the coefficients are:
$$ a_0 = 1, a_1 = 8, a_2 = 24, a_3 = 32, a_4 = 16 $$

Now, we can calculate the expression $$ a_0 - 2a_1 + 4a_2 - 8a_3 + 16a_4 $$ using these coefficients.
Calculate the value by following steps:
- step0: Calculate:
  $$1-2\times 8+4\times 24-8\times 32+16\times 16$$
- step1: Multiply the numbers:
  $$1-16+4\times 24-8\times 32+16\times 16$$
- step2: Multiply the numbers:
  $$1-16+96-8\times 32+16\times 16$$
- step3: Multiply the numbers:
  $$1-16+96-256+16\times 16$$
- step4: Multiply the numbers:
  $$1-16+96-256+256$$
- step5: Calculate:
  $$81$$
Calculate $$ a_0 - 2a_1 + 4a_2 - 8a_3 + 16a_4 $$ using the coefficients from the expansion of $$ (2x+1)^4 $$. Given: $$ a_0 = 1, \quad a_1 = 8, \quad a_2 = 24, \quad a_3 = 32, \quad a_4 = 16 $$ Plugging in the values: $$ 1 - 2(8) + 4(24) - 8(32) + 16(16) $$ $$ = 1 - 16 + 96 - 256 + 256 $$ $$ = 1 - 16 + 96 - 256 + 256 $$ $$ = 1 - 16 = -15 $$ $$ -15 + 96 = 81 $$ $$ 81 - 256 = -175 $$ $$ -175 + 256 = 81 $$ So, the result is 81.

\( 2 x+1)^{4}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4} \) रून, \( a_{0}-2 a_{1}+4 a_{2}-8 a_{3}+16 a \)

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