how that the coefficient of (r+1) th in the expansion of `(1+x)^(n+1)` is equal to the sum of the coefficients of the r th and (r+1) th term in the expansion of `(1+x)^n`

Answer

N/a

Explanation

in the expansion of `(1+x)^(n)`
`T_(r)=^(n)C_(r-1).(1)^(n-(r-1)).x^(r-1)`
` =^(n)C_(r-1).x^(r-1)`
Coefficient of rth term `=^(n)C_(r-1)`
`" and " T_(r+1) =^(n)C_(r)(1)^(n-r).x^(r) =^(n)C_(r)x^(r) `
`:.` Coefficient of (r+1)th term `=^(n+1)C_(r)`
Now the sum of the coefficient of rth and (r+1)th terms in the expansion of `(1+x)^(n)`
`=^(n)C_(r-1)+^(n)C_(r)=(|uln)/(|ul(r-1)|ul(n-r+1))+(|uln)/(|ulr|ul(n-r))`
`=(|uln)/(|ul(r-1)|ul(n-r)).[(1)/(n-r+1)+(1)/(r)]`
`=(|uln)/(|ul(r-1)|ul(n-r)).(r+n-r+1)/((n-r+1).r)`
`(|uln(n+1))/(|ulr|ul(n-r+1))=(|ul(n+1))/(|ulr|ul((n+1)-r))=^(n+1)C_(r)`
=Coefficient of (r+1)th term in the expansion of `(1+x)^(n+1)`. Hence Proved.