Above is question and its solution. I didn't get the first step itself. Explain in detail

Dear student,
given series is 1+x6+1+x7+.............+1+x15 ..............(1)
as we can see it is an G.P
and we know that a+ar+ar2+ar3+.......+arn-1=arn-1r-1
here a=1+x6 and r=1+x and n=10
now (1) will become 1+x61+x10-11+x-1
or 1+x16-1+x6x
to find coefficient of x6,we need to find the coefficient of x7 in 1+x16
we also know that coefficient of xr in 1+xn is nCr
so coefficient of ‚Äčx7 is 16C7=16C16-7=16C9     (since nCr=nCn-r)


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They have considered the expression to be a GP. The 'variable' of GP is took as (1+x) instead of x (in normal cases).
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Yup ! Reeshab is absolutely correct ...
But rishabh its not 'variable' , its 'common ratio'
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What are you looking for?