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QJEEM24Ma9891: JEE Main 2024 Mathematics - Class 11 JEEM24Ma9891 Solution: If , and , where , then equal to _________.

If S(x)=(1+x)+2(1+x)2+3(1+x)3+⋯+60(1+x)60,x≠0\mathrm{S}(x)=(1+x)+2(1+x)^2+3(1+x)^3+\cdots+60(1+x)^{60}, x \neq 0 , and (60)2 S(60)=a(b)b+b(60)^2 \mathrm{~S}(60)=\mathrm{a}(\mathrm{b})^{\mathrm{b}}+\mathrm{b} , where a,b∈Na, b \in N , then (a+b)(a+b) equal to _________.

Step-by-Step Verified Solution

JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Sequences and Series Question 15 English Explanation

amp;(−x)(S(x))=(1+x)[(1+x)60−1](1+x−1)−60(1+x)61amp;(−x)S(x)=(1+x)[(1+x)60−1]x−60(1+x)61amp;xS(x)=60(1+x)61−(1+x)[(1+x)60−1]x\begin{aligned}& (-x)(S(x))=\frac{(1+x)\left[(1+x)^{60}-1\right]}{(1+x-1)}-60(1+x)^{61} \\& (-x) S(x)=\frac{(1+x)\left[(1+x)^{60}-1\right]}{x}-60(1+x)^{61} \\& x S(x)=60(1+x)^{61}-\frac{(1+x)\left[(1+x)^{60}-1\right]}{x}\end{aligned}

Multiplying xx on both side,

x2S(x)=60x(1+x)61−(1+x)[(1+x)60−1]x^2 S(x)=60 x(1+x)^{61}-(1+x)\left[(1+x)^{60}-1\right]

Putting x=60x=60

amp;(60)2S(60)=60×60(61)61−(61)[6160−1]amp;=60×60(61)61−(61)⋅6160+61amp;=(61)61[60×60−1]+61amp;=(3600−1)⋅6161+61amp;a=3600−1,b=61⇒a+b=3660\begin{aligned}& (60)^2 S(60)=60 \times 60(61)^{61}-(61)\left[61^{60}-1\right] \\& =60 \times 60(61)^{61}-(61) \cdot 61^{60}+61 \\& =(61)^{61}[60 \times 60-1]+61 \\& =(3600-1) \cdot 61^{61}+61 \\& a=3600-1, \quad b=61 \Rightarrow a+b=3660\end{aligned}

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