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`        find the greatest term in the expansion of (7-5x)^11 when x=2/3`
one year ago

Arun
24740 Points
```							Dear Sara use the formula n+ 1 / 1 + |a/b| = 12 / (31/10) = 120/31 = 3._____ hence 4th term is greatest or you can also solve like thisBinomial expansion of (7 - 5 x)¹¹    when x = 2/3ie.,   (7 - 10/3)¹¹ = (21 - 10)¹¹ / 3¹¹We 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 10¹⁰We 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,  15So T0  T6, T6 > T8, T8 > T10So T4 is maximum.  It is   ¹¹C₄ 21⁷ 10⁴ / 3¹¹
```
one year ago
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