# Find numerically the greatest term in the expansion of (2 + 3x)9, where x = 32. - Mathematics

Sum

Find numerically the greatest term in the expansion of (2 + 3x)9, where x = 3/2.

#### Solution

We have (2 + 3x)9 = 2^9 (1 + (3x)/2)^9

Now, ("T"_(r + 1))/"T"_r = (2^9 [""^9"C"_r ((3x)/2)^r])/(2^9 [""^9"C"_(r - 1) ((3x)/2)^(r - 1)]

= (""^9"C"_r)/(""^9"C"_(r - 1)) |(3x)/2|

= 9/(r(9 - r)) * ((r - 1)(10 - r))/9 |(3x)/2|

= (10 - r)/r |(3x)/2|

= (10 - r)/r (9/4)

Since x = 3/2

Therefore, ("T"_(r + 1))/"T"_r ≥ 1

⇒ (90 - 9r)/(4r) ≥ 1

⇒ 90 – 9r ≥ 4r   ....(Why)

⇒ r ≤ 90/13

⇒ r ≤ 6 12/13

Thus the maximum value of r is 6

Therefore, the greatest term is Tr+1 = T7

Hence, T7 = 2^9 [""^9"C"_6 ((3x)/2)^6]

Where x = 3/2

= 2^9 * ""^9"C"_6 (9/4)^6

= 2^9 * (9 xx 8 xx 7)/(3 xx 2 xx 1) (3^12/2^12)

= (7 xx 3^13)/2

Concept: General and Middle Terms
Is there an error in this question or solution?

#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 8 Binomial Theorem
Solved Examples | Q 11 | Page 135

Share