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`        If (1+x)(1+x2)(1+x4).......(1+x128)=∑xr where r=0 and limit is n. So find the value of n.`
one year ago

1815 Points
```							since (1+x)(1+x2)(1+x4).......(1+x128)=∑xr where r=0 and limit is n, all we gotta do is to find the degree of the polynomial on LHS, which is equal to 1+2+4+....128this is a GP with common ratio 2, so sum is:S=1*(2^8 – 1)/(2 – 1)so n=S=2^8 – 1n= 255
```
one year ago
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