binomial theorem :

S_{n}= \sum_{r=1}^{n}\frac{ {(n+r)}!}{n!.r!}
n!S_n=\sum_{r=1}^{n}{\frac{(n+r)!}{r!}}
T_{r+1}= \frac{ {(n+r+1)}(n+r)!}{.(r+1)r!}
T_{r+1}= \frac{ {(n+r+1)}T_{r}}{(r+1)}
(r+1)T_{r+1}-rT_{r} = (n+1)T_{r}
(2)T_{2}-T_{1} = (n+1)T_{1}
(3)T_{3}-2T_{2} = (n+1)T_{2}
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(n+1)T_{n+1}-nT_{n} = (n+1)T_{n}
now we will observe everything except (n+1)T_n+1-T₁ gets cancelled
(n+1)S=(n+1)T_n+1-(n+1)
S=²ⁿ⁺¹C_n+1
sorry guyz i did a silly mistake
this giving the right answer
²ⁿ⁺¹C_n+1
but the solution in adg is one step.......
it is coefficient of xⁿ in the expansion
(1+x)ⁿ +(1+x)ⁿ⁺¹ +(1+x)ⁿ⁺² +............(1+x)²ⁿ
taking g.p summation the sum is finished

whats wrong then ??