**The sum of the coefficients of the first three terms in the expansion of (x-3/x ^{2})^{m} , x is not equal to 0,m being a natural number, is 559. Find the term of the wxpansion containing x^{3} **. (NCERT PG 174 EXAMPLE NO 16). The steps in the NCERTbook are not clear..

The coefficients of the first three terms of are * ^{m}*C

_{0}, (–3)

*C*

^{m}_{1}and (-3)

^{2}

*C*

^{m}_{2}.

Therefore, by the given condition, we have* ^{m}*C

_{0}–3

*C*

^{m}_{1}+ 9

*C*

^{m}_{2}= 559

since *m* is a natural number

⇒ *m* = 12

Now, T_{r}_{ + 1} = ^{12}C_{r} x^{12 – }* ^{r}* = =

^{12}C

*(– 3)*

_{r}*.*

^{r}*x*

^{12 – 3}

^{r}Since we need the term containing *x*^{3}

⇒ 12 - 3r = 3

⇒ 3*r *= 12 - 3 = 9

⇒ *r* = 3

Hence, the required term is ^{12}C_{3} (– 3)^{3} *x*^{3 }= = – 5940 *x*^{3}.

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