** Algebra > Binomial Theorem >solved Examples**

5. Find the coefficient of x^{44} in the binomial expansion
(2x^{3} - 3/_{x²})^{18}

*Solution:*

- We use the general term

- We must collect all the powers of x in the given binomial expansion

- We must set the powers of x in step 2 to 44. This gives us the value of r.

- We use r value obtained in the general term to find the coefficient of x
^{44}

T_{r + 1} = ^{n}c_{r} .x ^{n – r} . y^{r}

= ^{18}c_{r} . (2x^{3})^{ 18 – r} .(-3/_{x²} )^{r}

= ^{18}c_{r} . (2)^{ 18 – r} . (-3)^{r} . (x^{3})^{ 18– r} .1/_{x²} )^{r}

= ^{18}c_{r} . (2)^{ 18 – r} . (-3)^{r} . (x)^{54 – 3r} .(x)^{-2r}

= ^{18}c_{r} . (2)^{ 18 – r} . (-3)^{r} . (x)^{54 – 5r} ……………………(1)

Since we must find the coefficient of x^{44}, we must equate the power of x that is

54 – 5r in (1) to 44.

54 – 5r = 44

5r = 10, r = 2.

Therefore it is the third term, T_{3} that contains the coefficient of x^{44} in the given binomial expansion.

Now we will write this value of r, that is 2, in (1) to write the coefficient of x^{44}

^{18}c_{2} . (2)^{18 – 2 . (-3)2}

= 1377.(2)^{16}

6. Find the term independent of x in (2x^{3} + 3/_{x²} )^{10}

*Solution:*

Remember our discussion of independent term of x in a binomial expansion in formula 8 above earlier. Recall the steps as follows:

1. Use the general term first

2. Collect all the powers of x in the given binomial expansion.

3. Term independent of x means the term does not contain x. For the term to be without x or independent of x, the power of x should be equated to 0, since x^{0} = 1 and thus x
vanishes in the term into 1

T_{r + 1} = ^{n}c_{r} .x^{ n – r} . y^{ r}

= ^{10}c_{r} . (2x^{3})^{ 10 – r} . (3/_{x²} )^{r}

= ^{10}c_{r} (2)^{10 – r} . (x^{3})^{10 – r} (3)^{r}. (1/_{x²} )^{r}

= ^{10}c_{r} (2)^{10 – r} . (x)^{30 – 3r} . (3)^{r} . (x)^{ -2r}

= ^{10}c_{r} (2)^{10 – r} . (3)^{r} . (x)^{30 – 5r} ………………..(2)

Now set the power of x to 0 to make the term independent of x

30 – 5r = 0

r = 6.

Thus, the term independent of x or not having x is

T_{r + 1} = T_{6 + 1} = T_{7}

Hence, the seventh term T_{7} does not contain x

Lets write 6 in r in (2) to write this independent term of x

= ^{10}c_{6} (2)^{10 –6}. (3)^{6} . (x)^{30 – 5(6)}

= ^{10}c_{6} (2)^{10 –6}. (3)^{6} . x^{0}

= ^{10}c_{6} (2)^{4}. (3)^{6}

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