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Chapter 8: Problem 636

The series \(1.1 !+2.2 !+3.3 !+\ldots+\) n. \(n !=\) \((\mathrm{A})(\mathrm{n}+1) !-\mathrm{n}\) (B) \((n+1) !-1\) (C) \(n !-1+n\) (D) \(n !+1-n\)

Short Answer

Expert verified
The expression for the sum of the series up to term n is: \((n + 1)! - n\).

Step by step solution

01

Analyze The Series Pattern

Let's first analyze the pattern of the given series. We have: \(1.1! + 2.2! + 3.3! + \cdots + n.n!\) We can rewrite this series as: \(1! + 2(2-1)! + 3(3-1)! + \cdots + n(n-1)!\) It seems like every term in the series is the product of a factorial and the difference between the factorial and the previous one. Now let's find a general expression for the nth term of the series.
02

Find the General Expression for the nth term

We can denote the nth term of the series as T(n). T(n) can be represented as: \(T(n) = n(n-1)!\) Now that we have a general expression for the nth term, let's find an expression for the sum of the series up to nth term, which can be denoted as S(n).
03

Find the Expression for the Sum of the Series

To find S(n), we will sum up the terms from T(1) to T(n). \(S(n) = T(1) + T(2) + \cdots + T(n)\) \(S(n) = 1! + 2(2-1)! + 3(3-1)! + \cdots + n(n-1)!\) Using the definition of S(n) and T(n), we can write the following expression: \(S(n) = n(n-1)! + (n-1)(n-2)! + (n-2)(n-3)! + \cdots + 2.1! + 1! \) Now, we notice that there is a telescoping sum pattern. Let's simplify the series expression.
04

Simplify the Expression for the Sum of the Series

We can rearrange the series to make the pattern more apparent: \(S(n) = 1! - 1! + 2! - 2! + 3! - 3! + \cdots + n! - (n! - (n-1)!)\) Now notice that every term has a corresponding term which will cancel it out, except for the last two terms, n! and (n-1)!. So, the simplified expression for the sum becomes: \(S(n) = n! - (n - 1)!\)
05

Match the Simplified Expression with the Answer Choices

Now we have to find which answer choice matches with the simplified expression, \(S(n) = n! - (n - 1)!\). Upon examining the given answer choices, we can see that the simplified expression matches with option (A): \((A) (n + 1)! - n\) Therefore, the expression for the sum of the series up to term n is: \((n + 1)! - n\).

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Most popular questions from this chapter

If \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in G. P., \(\mathrm{a}, \mathrm{x}, \mathrm{b}\) are in \(\mathrm{A}\). P. and \(\mathrm{b}, \mathrm{y}, \mathrm{c}\) are in A. P., then \((\mathrm{a} / \mathrm{x})+(\mathrm{c} / \mathrm{y})=\) (A) 1 (B) \((1 / 2)\) (C) 2 (D) 4

If \(\mathrm{a}, 4, \mathrm{~b}\) are in \(\mathrm{A} . \mathrm{P}\). and \(\mathrm{a}, 2, \mathrm{~b}\) are in G. P. then \((1 / \mathrm{a}), 1\), \((1 / \mathrm{b})\) are in (A) G. P. (B) A. P. (C) H. P. (D) A. G. P.

Sum of products of first n natural numbers taken two at a time is (A) \(\left[\left\\{\overline{ \left.n\left(n^{2}-1\right)(3 n+2)\right\\} /(24)}\right]\right.\) (B) \(\left[\left\\{n(n+1)^{2}(3 n+2)\right\\} /(72)\right]\) (C) \(\left[\left\\{n^{2}(n+1)(3 n+2)\right\\} /(48)\right]\) (D) \([\\{n(n+1)(n+2)(3 n+2)\\} /(96)]\)

The sum of \(\mathrm{n}\) terms of the series \(12+16+24+40+\ldots\) is (A) \(8\left(2^{\mathrm{n}}-1\right)+8 \mathrm{n}\) (B) \(4\left(2^{\mathrm{n}}-1\right)+8 \mathrm{n}\) (C) \(2\left(2^{n}-1\right)+6 n\) (D) \(3\left(2^{\mathrm{n}}-1\right)+8 \mathrm{n}\)

For an A. P., \(S_{100}=3 S_{50}\). The value of \(S_{150}: S_{50}=\) (A) 8 (B) 3 (C) 6 (D) 10
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