Find the 4th term from the end in the expansion of 4 - 5-52 - 8-

Let Tr+1 be the 4th term from the end of the given expression. Then, Tr+1 is (10 minus; 4 + 1)th term, i.e., 7th term, from the beginning. Thus, we have: T7= ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

General Term of Binomial Expansion

Find the 4th ...

Question

Find the 4th term from the end in the expansion of ${(\frac{4 x}{5} - \frac{5}{2 x})}^{8}$ .

Open in App

Solution

Let T_r₊₁ be the4th term from the end of the given expression.
Then,
T_r₊₁is (10 − 4 + 1)th term, i.e., 7th term, from the beginning.
Thus, we have:
$T_{7} = T_{6 + 1} = {}^{9}C_{6} {(\frac{4 x}{5})}^{9 - 6} {(\frac{5}{2 x})}^{6} = \frac{9 \times 8 \times 7}{3 \times 2} (\frac{64}{125} x^{3}) (\frac{125 \times 125}{64 x^{6}}) = \frac{10500}{x^{3}}$

Suggest Corrections

Similar questions

Q. Find the 7th term in the expansion of

{(\frac{4 x}{5} + \frac{5}{2 x})}^{8}

Q. The 4th term from end in the expansion of

{(\frac{3}{x^{2}} - \frac{x^{3}}{5})}^{8}

Find the 4th term from the end in the expansion of ${(\frac{x^{3}}{2} - \frac{2}{x^{2}})}^{9}$

Q. Find the

4^{t h}

term from the end in the expansion of

{(\frac{x^{3}}{2} - \frac{2}{x^{2}})}^{7}

Q. Find the 4th term from the end in the expansion of

{(\frac{3}{x^{2}} - \frac{x^{3}}{6})}^{7}

Join BYJU'S Learning Program

Find the 4th term from the end in the expansion of 4x5-52x8.

Let Tr+1 be the 4th term from the end of the given expression. Then, Tr+1 is (10 − 4 + 1)th term, i.e., 7th term, from the beginning. Thus, we have: T7=T6+1 =C69 4x59-652x6 =9×8×73×264125x3125×12564x6 =10500x3

Find the 4th term from the end in the expansion of ${(\frac{4 x}{5} - \frac{5}{2 x})}^{8}$ .