Question
The number of terms which are free from radical signs in the expansion of (y15+x110)55 is
Answer: Option B
:
B
In the expansion of(y15+x110)55, the general
term is
Tr+1 =55Cr(y15)55−r(x110)r =55Cry11−r5xr10.
This Tr+1 will be independent of radicals if the exponents r5 and r10 are integers, for
0≤ r≤ 55 which is possible only when
r = 0, 10, 20, 30, 40, 50.
∴ There are six terms viz. T1,T11,T21,T31,T41,T51
which are independent of radicals.
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:
B
In the expansion of(y15+x110)55, the general
term is
Tr+1 =55Cr(y15)55−r(x110)r =55Cry11−r5xr10.
This Tr+1 will be independent of radicals if the exponents r5 and r10 are integers, for
0≤ r≤ 55 which is possible only when
r = 0, 10, 20, 30, 40, 50.
∴ There are six terms viz. T1,T11,T21,T31,T41,T51
which are independent of radicals.
Was this answer helpful ?
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