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Binomial Theorem — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsBinomial Theorem3 Marks
Question
Using binomial theorem write down the expansions of the following:
$(1+2\text{x}-3\text{x}^2)^5$
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Answer

Let $y = 1 + 2x,$ then
$(1+2\text{x}-3\text{x}^2)^5=(\text{y}-3\text{x}^2)^5$
The expansion of $(x + y)^n$  has $n + 1$ terms so the expansion of $(y - 3x^2)^5$  has $6$ terms.
Using binomial theorem to expand, we get
$(\text{y}-3\text{x}^2)^5={^5\text{C}}_0\text{y}^5(3\text{x}^2)^0-{^5\text{C}}_1\text{y}^4(3\text{x}^2)^1+{^5\text{C}}_2\text{y}^3(3\text{x}^2)^2\\-{^5\text{C}}_3\text{y}^2(3\text{x}^2)^3+{^5\text{C}}_4\text{y}(3\text{x}^2)^4-{^5\text{C}}_5\text{y}^0(3\text{x}^2)^5$
$=\text{y}^5-5\text{y}^43\text{x}^2+10\text{y}^39\text{x}^4-10\text{y}^2(27\text{x}^6)+5\text{y}81\text{x}^8-243\text{x}^{10}$
Now,
$\text{y}^5=(1+2\text{x})^5={^5\text{C}}_0+{^5\text{C}}_1(2\text{x})^1+{^5\text{C}}_2(2\text{x})^2\\+{^5\text{C}}_3(2\text{x})^3+{^5\text{C}}_4(2\text{x})^4+{^5\text{C}}_5(2\text{x})^5$
$\text{y}^4=(1+2\text{x})^4={^4\text{C}}_0+{^4\text{C}}_1(2\text{x})^1 +{^4\text{C}}_2(2\text{x})^2+{^4\text{C}}_3(2\text{x})^3+{^4\text{C}}_4(2\text{x})^4$
$\text{y}^3=(1+2\text{x})^3={^3\text{C}}_0+{^3\text{C}}_1(2\text{x}) +{^3\text{C}}_2(2\text{x})^2+{^3\text{C}}_3(2\text{x})^3$
$\text{y}^2=(1+2\text{x})^2={^2\text{C}}_0+{^2\text{C}}_1(2\text{x}) +{^2\text{C}}_2(2\text{x})^2$
$\text{y}=(1+2\text{x})$
Substituting the valus of powers of y in the equation above we get,
$\big(1+2\text{x}-3\text{x}^2\big)^5$$=\big[{^5\text{C}}_0+{^5\text{C}}_1(2\text{x})^1+{^5\text{C}}_2(2\text{x})^2+{^5\text{C}}_3(2\text{x})^3+{^5\text{C}}_4(2\text{x})^4+{^5\text{C}}_5(2\text{x})^5\big]$
$-15\text{x}^2\big[{^4\text{C}}_0+{^4\text{C}}_1(2\text{x})^1+{^4\text{C}}_2(2\text{x})^2+{^4\text{C}}_3(2\text{x})^3+{^4\text{C}}_4(2\text{x})^4\big]\\+90\text{x}^4\big[{^3\text{C}}_0+{^3\text{C}}_1(2\text{x})+{^3\text{C}}_2(2\text{x})^2+{^3\text{C}}_3(2\text{x})^3\big]-270\text{x}^6$$\big[{^2\text{C}}_0+{^2\text{C}}_1(2\text{x})+{^2\text{C}}_2(2\text{x})^2+5\times81\text{x}^8(1+2\text{x})-243\text{x}^{10}\big]$
$=10+10\text{x}+10\times4\text{x}^2+10\times8\text{x}^3+5\times16\text{x}^4\\+32\text{x}^5-15\text{x}^2-120\text{x}^3-180\text{x}^4+480\text{x}^5-240\text{x}^6+90\text{x}^4+540\text{x}^5\\+32\text{x}^5-15\text{x}^2-120\text{x}^3-180\text{x}^4+480\text{x}^5-240\text{x}^6+90\text{x}^4+540\text{x}^5\\+1080\text{x}^6+720\text{x}^7-270\text{x}^6-1080\text{x}^7-1080\text{x}^8+405\text{x}^8+810\text{x}^9-243\text{x}^{10}$
$=1+10\text{x}+25\text{x}^2-40\text{x}^3-190\text{x}^4+92\text{x}^5+570\text{x}^6\\-360\text{x}^7-675\text{x}^8+810\text{x}^9-243\text{x}^{10}$

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