Find the value of (7)/(2×5×9)+(7)/(5×9×12)+(7)/(9×12 ×16)+(7)/(12 ×16 ×19)+cdots+(7)/(30 ×33 ×37)

Recognize Pattern and Identify: Recognize the pattern in the series and identify the general term. The series is a sequence of terms of the form (7)/(a×(a+3)×(a+7)), where a starts at 2 and increases by 3 for each term.Write General Term: Write down the general term of the series. The nth term of the series can be written as T_n = (7)/(a_n×(a_n+3)×(a_n+7)), where a_n = 2 + 3(n-1).Simplify Expression for a_n: Simplify the expression for a_n. a_n = 2 + 3(n-1) = 2 + 3n - 3 = 3n - 1.Substitute into General Term: Substitute a_n into the general term. T_n = (7)/(3n - 1)×(3n + 2)×(3n + 6).Split into Partial Fractions: Notice that the general term can be split into partial fractions. We can express T_n as a sum of simpler fractions, A/(3n - 1) + B/(3n + 2) + C/(3n + 6), where A, B, and C are constants to be determined.Set up Equation for Decomposition: Set up the equation for partial fraction decomposition. 7 = A(3n + 2)(3n + 6) + B(3n - 1)(3n + 6) + C(3n - 1)(3n + 2).Solve for Constants: Solve for A, B, and C by plugging in convenient values for n. For n = 1/3, the first term vanishes, and we can solve for B. For n = -2/3, the second term vanishes, and we can solve for A. For n = -6/3, the third term vanishes, and we can solve for C.Plug in n = 1/3: Plug in n = 1/3 to solve for B. 7 = B(-1)(4) => B = -7/4.Plug in n = -2/3: Plug in n = -2/3 to solve for A. 7 = A(0)(2) => This does not work to solve for A, as the product is zero. We need to choose a different approach to find A and C.

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Find the value of
(7)/(2×5×9)+(7)/(5×9×12)+(7)/(9×12 ×16)+(7)/(12 ×16 ×19)+cdots+(7)/(30 ×33 ×37)

Find the value of $\frac{7}{2 \times 5 \times 9}+\frac{7}{5 \times 9 \times 12}+\frac{7}{9 \times 12 \times 16}+\frac{7}{12 \times 16 \times 19}+\cdots+\frac{7}{30 \times 33 \times 37}$

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Calculus

Evaluate definite integrals using the chain rule

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Q. Find the value of

\frac{7}{2 \times 5 \times 9}+\frac{7}{5 \times 9 \times 12}+\frac{7}{9 \times 12 \times 16}+\frac{7}{12 \times 16 \times 19}+\cdots+\frac{7}{30 \times 33 \times 37}

Recognize Pattern and Identify: Recognize the pattern in the series and identify the general term. $\newline$ The series is a sequence of terms of the form $\frac{7}{a\times(a+3)\times(a+7)}$ , where $a$ starts at $2$ and increases by $3$ for each term.
Write General Term: Write down the general term of the series. $\newline$ The $n$ th term of the series can be written as $T_n = \frac{7}{a_n\times(a_n+3)\times(a_n+7)}$ , where $a_n = 2 + 3(n-1)$ .
Simplify Expression for $a_n$ : Simplify the expression for $a_n$ . $\newline$ $a_n = 2 + 3(n-1) = 2 + 3n - 3 = 3n - 1.$
Substitute into General Term: Substitute $a_n$ into the general term. $\newline$ $T_n = \frac{7}{3n - 1}\times(3n + 2)\times(3n + 6).$
Split into Partial Fractions: Notice that the general term can be split into partial fractions. We can express $T_n$ as a sum of simpler fractions, $\frac{A}{3n - 1} + \frac{B}{3n + 2} + \frac{C}{3n + 6}$ , where $A$ , $B$ , and $C$ are constants to be determined.
Set up Equation for Decomposition: Set up the equation for partial fraction decomposition. $7 = A(3n + 2)(3n + 6) + B(3n - 1)(3n + 6) + C(3n - 1)(3n + 2)$ .
Solve for Constants: Solve for $A$ , $B$ , and $C$ by plugging in convenient values for $n$ . For $n = \frac{1}{3}$ , the first term vanishes, and we can solve for $B$ . For $n = -\frac{2}{3}$ , the second term vanishes, and we can solve for $A$ . For $n = -\frac{6}{3}$ , the third term vanishes, and we can solve for $C$ .
Plug in $n = \frac{1}{3}$ : Plug in $n = \frac{1}{3}$ to solve for $B$ . $7 = B(-1)(4) \Rightarrow B = -\frac{7}{4}$ .
Plug in $n = -\frac{2}{3}$ : Plug in $n = -\frac{2}{3}$ to solve for $A$ . $7 = A(0)(2) \Rightarrow$ This does not work to solve for $A$ , as the product is zero. We need to choose a different approach to find $A$ and $C$ .