Find the 7 th term in the expansion of 4 x -5-5-2 x 8

T_N =T_r+1=^nC_rx^n ry^r N=7, r=6, n=8, x =4x/5, y =5/2x T_7 =T_6+1=^8C_6 ( 4x/5 )^2 ( 5/2x )^6 =28 ×r^2/5^2× x^4 ×5^6/2^6 × x^6 =28/4×5^4/x^4=7 × 5 × 125/x^4 ...

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Byju's Answer

Standard XII

Mathematics

Index of (r+1)th Term from End When Counted from Beginning

Find the 7 th...

Question

Find the 7th term in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x})}^{8}$

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Solution

$T_{N} = T_{r + 1} =^{n} C_{r} x^{n - r} y^{r}$

$N = 7, r = 6, n = 8, x = \frac{4 x}{5}, y = \frac{5}{2 x}$

$T_{7} = T_{6 + 1} =^{8} C_{6} {(\frac{4 x}{5})}^{2} {(\frac{5}{2 x})}^{6}$

$= 28 \times \frac{r^{2}}{5^{2}} \times x^{4} \times \frac{5^{6}}{2^{6} \times x^{6}}$

$= \frac{28}{4} \times \frac{5^{4}}{x^{4}} = \frac{7 \times 5 \times 125}{x^{4}}$

$= \frac{4375}{x^{4}}$

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Find the 7th term in the expansion of (4x5+52x)8

TN=Tr+1=nCrxn−ryr N=7,r=6,n=8,x=4x5,y=52x T7=T6+1=8C6(4x5)2(52x)6 =28×r252×x4×5626×x6 =284×54x4=7×5×125x4 =4375x4

Find the 7th term in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x})}^{8}$