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Let $c_0, c_1, c_2, \ldots, c_n$ be the binomial coefficients in the expansion of $(1+x)^n$.
If $S_{n+1}=5 \cdot c_0+8 \cdot c_1+11 \cdot c_2+\ldots .(n+1)$ terms, then $S_{11}=$
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If $S_{n+1}=5 \cdot c_0+8 \cdot c_1+11 \cdot c_2+\ldots .(n+1)$ terms, then $S_{11}=$
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Verified Answer
The correct answer is:
$20480$
$S_{n+1}=5 C_0+8 C_1+11 C_2+\ldots .+(n+1)$ terms
$$
\begin{aligned}
& 5,8,11, \ldots .(n+1) \text { terms } \\
& T_{n+1}=5+(n+1-1) 3=5+3 n \\
& \therefore \quad T_{11}=5+3(10)=35 \\
& S_{11}=5 C_0+8 C_1+11 C_2+\ldots .+32 C_9+35 C_{10} \\
& \text { or, } S_{11}=35 C_{10}+32 C_9+\ldots . .+8 C_1+5 C_0
\end{aligned}
$$
Adding both equation,
$$
\begin{array}{ll}
\therefore \quad & 2 S_{11}=40 C_0+40 C_1+\ldots . .+40 C_{n-1}+40 C_{10} \\
& 2 \cdot S_{11}=2 \cdot 20\left(C_0+C_1+\ldots . .+C_{10}\right) \\
\therefore \quad & S_{11}=20 \times 2^{10}=20480 .
\end{array}
$$
$$
\begin{aligned}
& 5,8,11, \ldots .(n+1) \text { terms } \\
& T_{n+1}=5+(n+1-1) 3=5+3 n \\
& \therefore \quad T_{11}=5+3(10)=35 \\
& S_{11}=5 C_0+8 C_1+11 C_2+\ldots .+32 C_9+35 C_{10} \\
& \text { or, } S_{11}=35 C_{10}+32 C_9+\ldots . .+8 C_1+5 C_0
\end{aligned}
$$
Adding both equation,
$$
\begin{array}{ll}
\therefore \quad & 2 S_{11}=40 C_0+40 C_1+\ldots . .+40 C_{n-1}+40 C_{10} \\
& 2 \cdot S_{11}=2 \cdot 20\left(C_0+C_1+\ldots . .+C_{10}\right) \\
\therefore \quad & S_{11}=20 \times 2^{10}=20480 .
\end{array}
$$
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