Question

Find the greatest term in the expansion of $( 7 - 5x )^{11}$ when x = $\frac{2}{3}$

Answer 1

Binomial expansion of (7 - 5 x)¹¹    when x = 2/3
ie.,   (7 - 10/3)¹¹ = (21 - 10)¹¹ / 3¹¹

We consider the binomial expansion of (21 - 10)¹¹.

In the expansion alternate terms are negative. So we take only positive terms.

21¹¹,  ¹¹C₂ 21⁹ 10²,  ¹¹C₄ 21⁷ 10⁴,  ¹¹C₆ 21⁵  10⁶,  ¹¹C₈ 21³ 10⁸, ¹¹C₁₀ 21 10¹⁰

We take ratios of T0/T2, T2/T4, T4 / T6 ... Then we find that 21²/10² is there in every ratio. Its value is 4.41.  These ratios are:

4.41/55,  4.41/6 ,  4.41*5/7,  14/5,  15

So T0 < T2 < T4 .   T4 > T6, T6 > T8, T8 > T10

So T4 is maximum.  It is   ¹¹C₄ 21⁷ 10⁴ / 3¹¹

Answer 2

Binomial expansion of (7 - 5 x)¹¹    when x = 2/3

ie.,   (7 - 10/3)¹¹ = (21 - 10)¹¹ / 3¹¹

We consider the binomial expansion of (21 - 10)¹¹.

In the expansion alternate terms are negative. So we take only positive terms.

21¹¹,  ¹¹C₂ 21⁹ 10²,  ¹¹C₄ 21⁷ 10⁴,  ¹¹C₆ 21⁵  10⁶,  ¹¹C₈ 21³ 10⁸, ¹¹C₁₀ 21 10¹⁰

We take ratios of T0/T2, T2/T4, T4 / T6 ... Then we find that 21²/10² is there in every ratio. Its value is 4.41.  These ratios are:

4.41/55,  4.41/6 ,  4.41*5/7,  14/5,  15

So T0 < T2 < T4 .   T4 > T6, T6 > T8, T8 > T10

So T4 is maximum.  It is   ¹¹C₄ 21⁷ 10⁴ / 3¹¹