The sum of middle terms in the expansion of 2 a - a 2-49A- 63-32 a148-aB- 63-32 a138-aC- 63-32 a148-aD- 63-32 a138-a

The correct option is D Middle terms are T_9+1/2=T_5 and T_9+3/2=T_6 ∴ S=^9C_4(2a)^9 4( a^2/4 )^4+^9C_5(2a)^9 5( a^2/4 )^5 =126× 32a^5×a^3/256 126×16a^4×a^10/ ...

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Byju's Answer

Standard XI

Mathematics

Middle Terms

The sum of mi...

Question

The sum of middle terms in the expansion of ${(2 a - \frac{a^{2}}{4})}^{9}$ is

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Solution

The correct option is D
$M i d d l e t e r m s a r e T_{\frac{9 + 1}{2}} = T_{5} a n d T_{\frac{9 + 3}{2}} = T_{6} ∴ S =^{9} C_{4} (2 a)^{9 - 4} {(- \frac{a^{2}}{4})}^{4} +^{9} C_{5} (2 a)^{9 - 5} {(- \frac{a^{2}}{4})}^{5} = 126 \times 32 a^{5} \times \frac{a^{3}}{256} - 126 \times 16 a^{4} \times \frac{a^{10}}{1024} = \frac{63}{4} a^{13} - \frac{63}{32} a^{14} = \frac{63}{32} (8 - a) a^{13}$

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The sum of middle terms in the expansion of (2a−a24)9 is

The correct option is D Middle terms are T9+12=T5 and T9+32=T6∴ S=9C4(2a)9−4(−a24)4+9C5(2a)9−5(−a24)5=126×32a5×a3256−126×16a4×a101024=634a13−6332a14=6332(8−a)a13

The sum of middle terms in the expansion of ${(2 a - \frac{a^{2}}{4})}^{9}$ is