The sum of the coefficients of the first three terms in the expansion of (x-3/x2)m , x is not equal - Maths - Binomial Theorem

Question

The sum of the coefficients of the first three terms in
the expansion of (x-3/x2)m , x is not equal
to 0,m being a natural number, is 559 Find the term of the
wxpansion containing x3 (NCERT PG 174 EXAMPLE
NO 16) The steps in the NCERTbook are not clear

Aakash Sharma · Accepted Answer

The coefficients of the first three terms of ${(x - \frac{3}{x^{2}})}^{m}$ are ^mC₀, (–3) ^mC₁and (-3)² ^mC₂.

Therefore, by the given condition, we have
^mC₀ –3 ^mC₁+ 9 ^mC₂ = 559

$\Rightarrow \frac{m!}{0! (m - 0)!} - 3 \frac{m!}{1! (m - 1)!} + 9 \frac{m!}{2! (m - 2)!} = 559 \Rightarrow 1 - 3 \frac{m (m - 1)!}{(m - 1)!} + 9 \frac{m (m - 1) (m - 2)}{2 (m - 2)!} = 559 \Rightarrow 1 - 3 m + \frac{9 m (m - 1)}{2} = 559 \Rightarrow \frac{2 - 6 m + 9 m^{2} - 9 m}{2} = 559 \Rightarrow 9 m^{2} - 15 m + 2 = 559 \times 2 = 1118 \Rightarrow 9 m^{2} - 15 m + 2 - 1118 = 0 \Rightarrow 9 m^{2} - 15 m - 1116 = 0 \Rightarrow 3 (3 m^{2} - 5 m - 372) = 0 \Rightarrow 3 m^{2} - 5 m - 372 = 0 \Rightarrow 3 m^{2} - 36 m + 31 m - 372 = 0 \Rightarrow 3 m (m - 12) + 31 (m - 12) = 0 \Rightarrow (3 m + 31) (m - 12) = 0 \Rightarrow m = - \frac{31}{3}, 12$

since m is a natural number

⇒ m = 12

Now, T_r_{+ 1} = ¹²C_r x^{12 –}^r = = ¹²C_r (– 3)^r . x^{12 – 3}^r

Since we need the term containing x³

⇒ 12 - 3r = 3

⇒ 3r = 12 - 3 = 9

⇒ r = 3

Hence, the required term is ¹²C₃ (– 3)³ x³= $\frac{12!}{3! (12 - 3)!} \times (- 27) x^{3} = \frac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1 \times 9!} \times (- 27 x^{3}) = - 27 \times 220 \times x^{3}$ = – 5940 x³.

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The sum of the coefficients of the first three terms in the expansion of (x-3/x2)m , x is not equal to 0,m being a natural number, is 559. Find the term of the wxpansion containing x3 . (NCERT PG 174 EXAMPLE NO 16). The steps in the NCERTbook are not clear..

The sum of the coefficients of the first three terms in the expansion of (x-3/x²)^m , x is not equal to 0,m being a natural number, is 559. Find the term of the wxpansion containing x³ . (NCERT PG 174 EXAMPLE NO 16). The steps in the NCERTbook are not clear..