how do i do these types of q (1 Viewer)

yashbb · Aug 29, 2021

In the expansion of (2x3 + 3x−2)10, the general term is 10Ck (2x3)10−k (3x−2)k.
Show that this general term can be written as 10Ck 210 − k 3k x 30 − 5k.

cossine · Aug 29, 2021

Use ^ for superscript.

yashbb · Aug 29, 2021

cossine said:
Use ^ for superscript.

In the expansion of (2x3 + 3x−2)10, the general term is 10Ck (2x3)^(10−k )(3x−2)^k.
Show that this general term can be written as 10Ck 2^(10 − k) 3^k x 30 − 5^k.

Lith_30 · Aug 29, 2021

You gotta expand the brackets and simplify from there

${10\choose k}\cdot(2x^3)^{10-k}\cdot(3x^{-2})^k$
$={10\choose k}\cdot(2)^{10-k}\cdot(x^3)^{10-k}\cdot3^{k}\cdot(x^{-2})^k$
$={10\choose k}\cdot2^{10-k}\cdot{3^{k}}\cdot{x}^{30-3k}\cdot{x}^{-2k}$
$={10\choose k}\cdot2^{10-k}\cdot{3^{k}}\cdot{x}^{30-5k}$

yashbb · Aug 30, 2021

Lith_30 said:
You gotta expand the brackets and simplify from there

${10\choose k}\cdot(2x^3)^{10-k}\cdot(3x^{-2})^k$
$={10\choose k}\cdot(2)^{10-k}\cdot(x^3)^{10-k}\cdot3^{k}\cdot(x^{-2})^k$
$={10\choose k}\cdot2^{10-k}\cdot{3^{k}}\cdot{x}^{30-3k}\cdot{x}^{-2k}$
$={10\choose k}\cdot2^{10-k}\cdot{3^{k}}\cdot{x}^{30-5k}$

tysm that helps ty ty.

how do i do these types of q (1 Viewer)

yashbb

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cossine

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yashbb

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Lith_30

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yashbb

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