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Binomial Theorem — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSBinomial Theorem1 Mark
Question
Expand $\left( x ^ { 2 } + \frac { 3 } { x } \right) ^ { 4 }$,  x $\neq$ 0
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Answer

By using binomial theorem, we have
(a + b)n = nC0 an + nC1 an-1 b + nC2 an-2 b2 + ... + nCr an-r br + ... + nCn bn
$\therefore$ $\left( x ^ { 2 } + \frac { 3 } { x } \right) ^ { 4 }$ = 4C0 (x2)4 + 4C1 (x2)4-1$\left( \frac { 3 } { x } \right) ^ { 1 }$ + 4C2 (x2)4-2 $\left( \frac { 3 } { x } \right) ^ { 2 }$ + 4C3(x2)4-3 $\left( \frac { 3 } { x } \right) ^ { 3 }$+ 4C4(x2)4-4 $\left( \frac { 3 } { x } \right) ^ { 4 }$
= 1.x8 + 4 (x2)3 $\left( \frac { 3 } { x } \right)$ + 6 (x2)2 $\left( \frac { 3 } { x } \right) ^ { 2 }$ + 4 (x2) $\left( \frac { 3 } { x } \right) ^ { 3 }$ + 1. (x2)0 $\left( \frac { 3 } { x } \right) ^ { 4 }$
[$\because$ 4C0 = 1, 4C1 = 4, 4C2 = 6, 4C3 = 4 and 4C4 = 1]
= $x ^ { 8 } + 4 \cdot x ^ { 6 } \cdot \frac { 3 } { x } + 6 x ^ { 4 } \cdot \frac { 9 } { x ^ { 2 } } + 4 x ^ { 2 } \cdot \frac { 27 } { x ^ { 3 } } + 1 \cdot 1 \cdot \frac { 81 } { x ^ { 4 } }$
= x8 + 12x5 + 54x2 + $\frac { 108 } { x } + \frac { 81 } { x ^ { 4 } }$

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