prodlimits_{n = 1}^{10} {left( {frac{{left( { | vedclass

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(A) Let the general term be $a_n = \frac{6\sum_{i=0}^n i + 1}{6\sum_{j=0}^n (j-1) + 1}$.Using the sum formulas $\sum_{i=0}^n i = \frac{n(n+1)}{2}$ and

Vedclass · Accepted Answer

$331$

Vedclass · Answer

(A) Let the general term be $a_n = \frac{6\sum_{i=0}^n i + 1}{6\sum_{j=0}^n (j-1) + 1}$.Using the sum formulas $\sum_{i=0}^n i = \frac{n(n+1)}{2}$ and $\sum_{j=0}^n (j-1) = \frac{n(n+1)}{2} - (n+1) = \frac{(n+1)(n-2)}{2}$.The numerator is $6 \cdot \frac{n(n+1)}{2} + 1 = 3n^2 + 3n + 1$.The denominator is $6 \cdot \frac{(n+1)(n-2)}{2} + 1 = 3(n^2 - n - 2) + 1 = 3n^2 - 3n - 5$. Wait,let's re-evaluate the denominator sum: $\sum_{j=0}^n (j-1) = -1 + 0 + 1 + ... + (n-1) = \frac{n(n-1)}{2} - 1$. Actually,$\sum_{j=0}^n (j-1) = \sum_{j=1}^n j - \sum_{j=0}^n 1 = \frac{n(n+1)}{2} - (n+1) = \frac{(n+1)(n-2)}{2}$.Correcting the expression: The term is $\frac{3n^2+3n+1}{3n^2-3n+1}$.For $n=1$: $\frac{3+3+1}{3-3+1} = \frac{7}{1}$.For $n=2$: $\frac{12+6+1}{12-6+1} = \frac{19}{7}$.For $n=3$: $\frac{27+9+1}{27-9+1} = \frac{37}{19}$.This is a telescoping product: $\prod_{n=1}^{10} \frac{f(n)}{f(n-1)}$ where $f(n) = 3n^2+3n+1$.The product is $\frac{f(10)}{f(0)} = \frac{3(100)+3(10)+1}{3(0)+3(0)+1} = \frac{331}{1} = 331$.

$\prod\limits_{n = 1}^{10} {\left( {\frac{{\left( {6\sum\limits_{i = 0}^n i } \right) + 1}}{{\left( {6\sum\limits_{j = 0}^n {(j - 1)} } \right) + 1}}} \right)} $ is equal to:

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