Let $\Delta a=\left|\begin{array}{ccc}a-1 & n & 6 \ (a-1)^{2} & 2 n^{2} & 4 n-2 \ (a-1)^{3} & 3 n^

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Let $\Delta a=\left|\begin{array}{ccc}a-1 & n & 6 \ (a-1)^{2} & 2 n^{2} & 4 n-2 \ (a-1)^{3} & 3 n^{3} & 3 n^{2}-3 n\end{array}ight|$. The $\sum_{a=1}^{n} \Delta a=f(n)$, a polynomial in n, then degree of n is

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Since $C_{2}$ and $C_{3}$ are independent of $a$, summation runs on $C_{1}$.\ $	herefore \sum_{a=1}^{n} \Delta a=\left|\begin{array}{ccc}\sum_{1}^{n}(a-1) & n & 6 \ \sum_{1}^{n}(a-1)^{2} & 2 n^{2} & 4 n-2 \ \sum_{1}^{n}(a-1)^{3} & 3 n^{3} & 3 n^{2}-3 n\end{array}ight|$ $$ =\left|\begin{array}{ccc} \frac{1}{2}(n-1) n & n & 6 \ \frac{1}{6}(n-1) n(2 n-1) & 2 n^{2} & 4 n-2 \ \frac{1}{4}(n-1)^{2} n^{2} & 3 n^{3} & 3 n^{2}-3 n \end{array}ight| $$ Taking $\frac{1}{12} n(n-1)$ as common factor from $C_{1}$ and $n$ as common factor from $C_{2}$ we get, $$ \sum \Delta a=\frac{1}{12} n^{2}(n-1)\left|\begin{array}{ccc} 6 & 1 & 6 \ 2(2 n-1) & 2 n & 2(2 n-1) \ 3 n(n-1) & 3 n^{2} & 3 n(n-1) \end{array}ight|=0 $$ (Since $C_{1}$ and $C_{3}$ are identical) $$ 	ext { 188. Let } \Delta=\left|\begin{array}{lll} { }^{x} C_{r} & { }^{x} C_{r+1} & { }^{x} C_{r+2} \ { }^{y} C_{r} & { }^{y} C_{r+1} & { }^{y} C_{r+2} \ { }^{z} C_{r} & { }^{z} C_{r+1} & { }^{z} C_{r+2} \end{array}ight| $$ Operating $C_{3} ightarrow C_{3}+C_{2}$, then $C_{2} ightarrow C_{2}+C_{1}$ and using ${ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}$ $$ \Delta=\left|\begin{array}{lll} { }^{x} C_{r} & { }^{x+1} C_{r+1} & { }^{x+1} C_{r+2} \ { }^{y} C_{r} & { }^{y+1} C_{r+1} & { }^{y+1} C_{r+2} \ { }^{z} C_{r} & { }^{z+1} C_{r+1} & { }^{z+1} C_{r+2} \end{array}ight| $$ Again operating $C_{3} ightarrow C_{3}+C_{2}$, we get $$ \Delta=\left|\begin{array}{lll} { }^{x} C_{r} & { }^{x+1} C_{r+1} & { }^{x+2} C_{r+2} \ { }^{y} C_{r} & { }^{y+1} C_{r+1} & { }^{y+2} C_{r+2} \ { }^{2} C_{r} & { }^{z+1} C_{r+1} & { }^{z+2} C_{r+2} \end{array}ight| $$