If a=1+2+4+cdots up to n terms,b=1+3+9+cdots | vedclass

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(C) $a = 1 + 2 + 4 + \cdots + 2^{n-1} = \frac{1(2^n - 1)}{2 - 1} = 2^n - 1$.$b = 1 + 3 + 9 + \cdots + 3^{n-1} = \frac{1(3^n - 1)}{3 - 1} = \frac{3^n -

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(C) $a = 1 + 2 + 4 + \cdots + 2^{n-1} = \frac{1(2^n - 1)}{2 - 1} = 2^n - 1$.$b = 1 + 3 + 9 + \cdots + 3^{n-1} = \frac{1(3^n - 1)}{3 - 1} = \frac{3^n - 1}{2} \Rightarrow 2b = 3^n - 1$.$c = 1 + 5 + 25 + \cdots + 5^{n-1} = \frac{1(5^n - 1)}{5 - 1} = \frac{5^n - 1}{4} \Rightarrow 4c = 5^n - 1$.Substituting these into the determinant:$\Delta = \left|\begin{array}{ccc} 2^n - 1 & 3^n - 1 & 5^n - 1 \ 2 & 2 & 2 \ 2^n & 3^n & 5^n \end{array}ight|$.Using the property of determinants,we can split the first row:$\Delta = \left|\begin{array}{ccc} 2^n & 3^n & 5^n \ 2 & 2 & 2 \ 2^n & 3^n & 5^n \end{array}ight| - \left|\begin{array}{ccc} 1 & 1 & 1 \ 2 & 2 & 2 \ 2^n & 3^n & 5^n \end{array}ight|$.In the first determinant,row $1$ and row $3$ are identical,so its value is $0$.In the second determinant,row $2$ is $2 	imes$ row $1$,so its value is $0$.Therefore,$\Delta = 0 - 0 = 0$.

If $a=1+2+4+\cdots$ up to $n$ terms,$b=1+3+9+\cdots$ up to $n$ terms and $c=1+5+25+\cdots$ up to $n$ terms,then $\Delta=\left|\begin{array}{ccc}a & 2b & 4c \\ 2 & 2 & 2 \\ 2^n & 3^n & 5^n\end{array}\right|=$

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