**Algebra > Binomial Theorem Examples**

**7. Find the greatest binomial coefficient in the binomial expansion (3x+2y) ^{12}.**

*Solution:*

In the given binomial expansion, the index n is 12, an even number. Therefore from our discussion in Greatest Binomial Coefficients , we see that the greatest binomial coefficient will be
^{n}c_{( n/2 + 1)}.

Therefore it is ^{12}c_{(12/2 + 1)} = ^{12}c_{(6+1)} = ^{12}c_{7}.

**8. Find the greatest binomial coefficient in the binomial expansion of
(2x – 4y) ^{11}**

*Solution:*

In the given binomial expansion, the index n is 11, an odd number. Therefore from our discussion in Greatest Binomial Coefficients, we see that there are two greatest binomial coefficients. And,they are:

^{n}c_{(n+1)/2} and ^{n}c_{( n + 3 )/2}.

Applying 11 in n:

^{11}c_{(11+1)/2} = ^{11}c_{6} and
^{11}c_{(11+3)/2} = ^{11}c_{7}

9. Find the value of

^{20}C_{1} + ^{20}C_{3} + ^{20}C_{5} + ^{20}C_{7} + ………….. + ^{20}C_{19}

*Solution:*

From properties of binomial coefficients , we know that

c_{0} + c_{2} + c_{4} +……….. = c_{1} + c_{3} + c_{5} + …………. = 2^{n – 1}

Since only sum of the odd binomial coefficients is required, we can apply the above formula:

^{20}C_{1} + ^{20}C_{3} + ^{20}C_{5} + ^{20}C_{7} + ………….. + ^{20}C_{19} = 2^{20 – 1}

**10. In the binomial expansion (2x ^{2} – 3/_{x})^{15}, the term which does not have x is ^{15}c_{10} . 3^{10} . K, where K is a real number. Find K**

*Solution.*

The term is independent of x ( since it does not have x ). Therefore, we will use general term to find r which is independent of x

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

= ^{15}c_{r} ( 2x^{2})^{ 15 – r} (^{ - 3} /_{ x})^{ r}

= ^{15}c_{r} (2)^{15 – r} ( - 3 )^{ r} (x^{2})^{ 15 – r} (1/_{x})^{r}

= ^{15}c_{r} (2)^{15 – r} ( - 3 )^{ r} (x^{30 – 2r})(x)^{- r}

= ^{15}c_{r} (2)^{15 – r} ( - 3 )^{ r} (x^{ 30 – 3r})……………………(3)

Since T_{r + 1} is independent of x, equate 30 – 3r to 0

30 – 3r = 0, 3r = 30, r = 10.

Since r = 10, it is the 11^{th} term that does not have x.

Plug in r = 10 in (3) above to find K.

= ^{15}c_{10} (2)^{15 – 10} ( - 3 )^{10}

= ^{15}c_{10} ( 3 )^{10} (2)^{5} ……………..(4)

Comparing the value in (4) above with ^{15}c_{10} .3^{10} . K given in the question, we observe that

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