In the expansion of ( 5 1 2 + 7 1 8 ) 1024 , the number of integral terms is Options: A . &nbsp 128 B . &nbsp 129 C . &nbsp 130 D . &nbsp 131

Answer: Option B : B Here n = 1024 = 2 10 , a power of 2, where as the power of 7 is 1 8 = 2 − 3 Now first term 1024 C 0 ( 5 1 2 ) 1024 = 5 512 (integer) And after 8 terms, the 9 t h term = 1024 C 8 ( 5 1 2 ) 1016 ( 7 1 8 ) 8 = an integer Again, 17 t h term = 1024 C 16 ( 5 1 2 ) 1008 ( 7 1 8 ) 16 = An integer. Continuing like this, we get an A.P. 1, 9, 17,..........., 1025, because 1025 t h term = the last term in the expansion = 1024 C 1024 ( 7 1 8 ) 1024 = 7 128 (an integer) If n is the number of terms of above A.P. we have 1025 = T n = 1 + (n - 1)8 ⇒ n = 129.

In the expansion of (512+718)1024, the number of integral terms is

Discussion Forum : Binomial Theorem

Question -

In the expansion of $(5^{\frac{1}{2}} + 7^{\frac{1}{8}})^{1024}$ , the number of integral terms is

Options:

A . 128

B . 129

C . 130

D . 131

Answer: Option B
:
B

Here n = 1024 = $2^{10}$ , a power of 2, where as the

power of 7 is $\frac{1}{8}$ = $2^{- 3}$

Now first term $^{1024} C_{0}$ ( $5^{\frac{1}{2}})^{1024}$ = $5^{512}$ (integer)

And after 8 terms, the $9^{t h}$ term = $^{1024} C_{8}$ ( $5^{\frac{1}{2}})^{1016}$ ( $7^{\frac{1}{8}})^{8}$ = an integer

Again, $17^{t h}$ term = $^{1024} C_{16}$ ( $5^{\frac{1}{2}})^{1008}$ ( $7^{\frac{1}{8}})^{16}$

= An integer.
Continuing like this, we get an A.P. 1, 9, 17,..........., 1025,

because $1025^{t h}$ term = the last term in the expansion

= $^{1024} C_{1024}$ ( $7^{\frac{1}{8}})^{1024}$ = $7^{128}$ (an integer)

If n is the number of terms of above A.P. we have

1025 = $T_{n}$ = 1 + (n - 1)8 ⇒ n = 129.

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