(A) The numerator is of the form $a^3 + b^3 + 3ab(a + b) = (a + b)^3$,where $a = 18$ and $b = 7$.Since $a + b = 18 + 7 = 25$,the numerator is $25^3$.T

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(A) The numerator is of the form $a^3 + b^3 + 3ab(a + b) = (a + b)^3$,where $a = 18$ and $b = 7$.Since $a + b = 18 + 7 = 25$,the numerator is $25^3$.The denominator is of the form $(x + y)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} y^k$.Here,$x = 3$ and $y = 2$,so the denominator is $(3 + 2)^6 = 5^6$.Since $5^6 = (5^2)^3 = 25^3$,the denominator is $25^3$.Therefore,the value of the expression is $\frac{25^3}{25^3} = 1$.

Class 11 Mathematics Binomial Theorem Expansion of binomial theorem

Medium

Find the value of $\frac{18^3 + 7^3 + 3 \times 18 \times 7 \times 25}{3^6 + 6 \times 243 \times 2 + 15 \times 81 \times 4 + 20 \times 27 \times 8 + 15 \times 9 \times 16 + 6 \times 3 \times 32 + 64}$

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A
$1$
B
$5$
C
$25$
D
$100$

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Binomial Theorem

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Expansion of binomial theorem

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IIT 1960 Paper All IIT years →

Expand using Binomial Theorem $\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \neq 0$.

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If the ratio of the coefficients of the third and fourth terms in the expansion of $(x - \frac{1}{2x})^n$ is $1 : 2$,then the value of $n$ is:

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$2{C_0} + \frac{2^2}{2}{C_1} + \frac{2^3}{3}{C_2} + \dots + \frac{2^{11}}{11}{C_{10}} = \dots$

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Find the value of $(a^{2}+\sqrt{a^{2}-1})^{4}+(a^{2}-\sqrt{a^{2}-1})^{4}$.

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Evaluate the expression: $\frac{1}{81^{n}} - {}^{2n}C_1 \frac{10}{81^{n}} + {}^{2n}C_2 \frac{10^2}{81^{n}} - \dots + \frac{10^{2n}}{81^{n}} = $

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Find the value of $\frac{18^3 + 7^3 + 3 \times 18 \times 7 \times 25}{3^6 + 6 \times 243 \times 2 + 15 \times 81 \times 4 + 20 \times 27 \times 8 + 15 \times 9 \times 16 + 6 \times 3 \times 32 + 64}$

Similar Questions

Expand using Binomial Theorem $\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \neq 0$.

If the ratio of the coefficients of the third and fourth terms in the expansion of $(x - \frac{1}{2x})^n$ is $1 : 2$,then the value of $n$ is:

$2{C_0} + \frac{2^2}{2}{C_1} + \frac{2^3}{3}{C_2} + \dots + \frac{2^{11}}{11}{C_{10}} = \dots$

Find the value of $(a^{2}+\sqrt{a^{2}-1})^{4}+(a^{2}-\sqrt{a^{2}-1})^{4}$.

Evaluate the expression: $\frac{1}{81^{n}} - {}^{2n}C_1 \frac{10}{81^{n}} + {}^{2n}C_2 \frac{10^2}{81^{n}} - \dots + \frac{10^{2n}}{81^{n}} = $

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