1no i saw it before you edited....
but still y do it by table form... u may miss some term...
i dunno if u are comfortable its ok :)
actually i feel confused with that method...
the number of terms of expansion with reational co-efficients in the expansion of (3√5+√3+z)6 is
a) 7 b) 6 c)8 d)9
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7 Answers
1there will be 7 rational terms
51/3 31/2 z
6 0 0
3 2 1
3 0 3
0 6 0
0 4 2
0 2 4
0 0 6
1(3√5+√3+z)6
= [(3√5+(√3+z)]6
tr+1 = 6Cr 56-r/3(√3+z)r [r<=6]
tk+1 = rCkzr-k3k/2 [k<=r]
now tr+1 is ratioal if r=0,3,6..
and tk+1 is raional for k=0,2,4,6
so, r k
0 0
3 0,2
6 0,2,4,6.
so 7 rational terms.
1
1@dimension,...okie..it depends.
@prophet sir, will u pls solve this one by multinomial..
i actually dunno it .. [2]
