If $1.3 .5+3.5 .7+5.7 .9+\ldots n$ terms $=n(n+1) f(n)-3 n$, then $f(1)=$

If $1.3 .5+3.5 .7+5.7 .9+\ldots n$ terms $=n(n+1) f(n)-3 n$, then $f(1)=$
  1. 9
  2. 11
  3. 12
  4. 8

Solution

$1 \cdot 3 \cdot 5+3 \cdot 5 \cdot 7+5 \cdot 7 \cdot 9+\ldots . n$ terms $\begin{aligned} & T_n=(2 n-1)(2 n+1)(2 n+3)=8 n^3+12 n^2-2 n-3 \\ & S_n=\Sigma T_n=8 \Sigma n^3+12 \Sigma n^2-2 \Sigma n-3 \Sigma 1\end{aligned}$
$\begin{aligned} & =\frac{8(n(n+1))^2}{4}+\frac{12 n(n+1)(2 n+1)}{6}-\frac{2 n(n+1)}{2}-3 n \\ & =2 n^4+8 n^3+7 n^2-2 n=n(n+1) f(n)-3 n\end{aligned}$
$\begin{aligned} & \Rightarrow f(x)=\frac{n\left(2 n^3+8 n^2+7 n+1\right)}{n(n+1)} \\ & \Rightarrow f(1)=\frac{2+8+7+1}{2}=9\end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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