Q.5. is equal to :

Q: What is the correct answer for JEE Main Mathematics Q5?

The correct answer is B. Sol. Let $\lambda = \sum_{n=1}^{10} \frac{528}{n(n+1)(n+2)}$Let $T_n = \frac{1}{n(n+1)(n+2)}$We can use partial fraction decomposition:$\frac{1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}$Multiplying by $n(n+1)(n+2)$: $1 = A(n+1)(n+2) + B(n)(n+2) + C(n)(n+1)$For $n=0$: $1 = A(1)(2) \Rightarrow A = 1/2$For $n=-1$: $1 = B(-1)(1) \Rightarrow B = -1$For $n=-2$: $1 = C(-2)(-1) \Rightarr

Q: Q.5. is equal to :

The correct answer is B. Sol. Let $\lambda = \sum_{n=1}^{10} \frac{528}{n(n+1)(n+2)}$Let $T_n = \frac{1}{n(n+1)(n+2)}$We can use partial fraction decomposition:$\frac{1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}$Multiplying by $n(n+1)(n+2)$: $1 = A(n+1)(n+2) + B(n)(n+2) + C(n)(n+1)$For $n=0$: $1 = A(1)(2) \Rightarrow A = 1/2$For $n=-1$: $1 = B(-1)(1) \Rightarrow B = -1$For $n=-2$: $1 = C(-2)(-1) \Rightarr

Q: What are the options for this JEE Main Mathematics question?

A. 65 · B. 130 · C. 220 · D. 440

Question

Q.5.  is equal to :

FirstInTest · Accepted Answer

Sol. Let $\lambda = \sum_{n=1}^{10} \frac{528}{n(n+1)(n+2)}$Let $T_n = \frac{1}{n(n+1)(n+2)}$We can use partial fraction decomposition:$\frac{1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}$Multiplying by $n(n+1)(n+2)$: $1 = A(n+1)(n+2) + B(n)(n+2) + C(n)(n+1)$For $n=0$: $1 = A(1)(2) \Rightarrow A = 1/2$For $n=-1$: $1 = B(-1)(1) \Rightarrow B = -1$For $n=-2$: $1 = C(-2)(-1) \Rightarrow C = 1/2$So, $T_n = \frac{1}{2n} - \frac{1}{n+1} + \frac{1}{2(n+2)}$This can be rewritten as: $T_n = \frac{1}{2} \left( \frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2} ight)$Or, $T_n = \frac{1}{2} \left( \left( \frac{1}{n} - \frac{1}{n+1} ight) - \left( \frac{1}{n+1} - \frac{1}{n+2} ight) ight)$This is a telescoping sum.Let's write out the terms for the sum $\sum_{n=1}^{10} T_n$:$\sum_{n=1}^{10} T_n = \frac{1}{2} \sum_{n=1}^{10} \left( \frac{1}{n} - \frac{1}{n+1} - \left( \frac{1}{n+1} - \frac{1}{n+2} ight) ight)$Let $f(n) = \frac{1}{n} - \frac{1}{n+1}$. Then $T_n = \frac{1}{2} (f(n) - f(n+1))$.$\sum_{n=1}^{10} T_n = \frac{1}{2} [ (f(1) - f(2)) + (f(2) - f(3)) + \dots + (f(10) - f(11)) ]$$= \frac{1}{2} [ f(1) - f(11) ]$$= \frac{1}{2} \left[ \left( \frac{1}{1} - \frac{1}{2} ight) - \left( \frac{1}{11} - \frac{1}{12} ight) ight]$$= \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{11 \cdot 12} ight]$$= \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{132} ight]$$= \frac{1}{2} \left[ \frac{66-1}{132} ight] = \frac{1}{2} \left[ \frac{65}{132} ight] = \frac{65}{264}$So, $\lambda = 528 	imes \frac{65}{264}$$\lambda = 2 	imes 65 = 130$

Answer

65

Answer

130

Answer

220

Answer

440

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