If S(x)=(1+x)+2(1+x)^2+3(1+x)^3+ +60(1+x)^60,x 0, and (60)^2S(60)=a(b)^b+b, where a,b N, t…

Question

If $S(x)=(1+x)+2(1+x{)}^{2}+3(1+x{)}^{3}+\cdots +60(1+x{)}^{60},x
e 0$, and $(60{)}^{2}S(60)=a(b{)}^{b}+b$, where $a,b\in N$, then $(a+b)$ equal to \_\_\_\_\_\_\_\_\_.

Answer: 3660

Accepted Answer

$\begin{gathered} (-x)(S(x))=\frac{(1+x)[(1+x{)}^{60}-1]}{(1+x-1)}-60(1+x{)}^{61} \ (-x)S(x)=\frac{(1+x)[(1+x{)}^{60}-1]}{x}-60(1+x{)}^{61} \ xS(x)=60(1+x{)}^{61}-\frac{(1+x)[(1+x{)}^{60}-1]}{x} \end{gathered}$

Multiplying $x$ on both side,

${x}^{2}S(x)=60x(1+x{)}^{61}-(1+x)[(1+x{)}^{60}-1]$

Putting $x=60$

$\begin{gathered} (60{)}^{2}S(60)=60	imes 60(61{)}^{61}-(61)[{61}^{60}-1] \ =60	imes 60(61{)}^{61}-(61)\cdot {61}^{60}+61 \ =(61{)}^{61}[60	imes 60-1]+61 \ =(3600-1)\cdot {61}^{61}+61 \ a=3600-1,\, b=61\Rightarrow a+b=3660 \end{gathered}$

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