The 6th term in the expansion of 2x2 - 13x210 is

We know that, (r+1)^th term in the expansion of (x+y)^n is T_r+1 = ^nC_r x^n ry^r Hence for (2x^2 13x^2)^10 T_6 = ^10C_5(2x^2)^5( 13x^2)^5 = 10!5!×5!×32 × ...

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Standard XII

Mathematics

General Term of Binomial Expansion

The 6th term ...

Question

The $6^{th}$ term in the expansion of ${(2 x^{2} - \frac{1}{3 x^{2}})}^{10}$ is

- \frac{35840}{9}

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\frac{560}{243}

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- \frac{896}{27}

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- \frac{764}{25}

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Solution

The correct option is C $- \frac{896}{27}$
We know that, $(r + 1)^{th}$ term in the expansion of $(x + y)^{n}$ is
$T_{r + 1} =^{n} C_{r} x^{n - r} y^{r}$
Hence for ${(2 x^{2} - \frac{1}{3 x^{2}})}^{10}$
$T_{6} =^{10} C_{5} (2 x^{2})^{5} {(- \frac{1}{3 x^{2}})}^{5}$
$= - \frac{10!}{5! \times 5!} \times 32 \times \frac{1}{243}$

$= - \frac{896}{27}$

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General Term of Binomial Expansion

Standard XII Mathematics

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The 6th term in the expansion of (2x2−13x2)10 is

The correct option is C −89627We know that, (r+1)th term in the expansion of (x+y)n is Tr+1= nCr xn−ryr Hence for (2x2−13x2)10 T6= 10C5(2x2)5(−13x2)5 =−10!5!×5!×32×1243 =−89627

The $6^{th}$ term in the expansion of ${(2 x^{2} - \frac{1}{3 x^{2}})}^{10}$ is