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Question: For any integer , the sum \[_{k = 1}^{k = n}{S_n}k(k + 2)\] is equal to 1) \[\dfrac{{n(n + 1)(n +...

For any integer , the sum k=1k=nSnk(k+2)_{k = 1}^{k = n}{S_n}k(k + 2) is equal to

  1. n(n+1)(n+2)2\dfrac{{n(n + 1)(n + 2)}}{2}
  2. n(n+1)(2n+1)6\dfrac{{n(n + 1)(2n + 1)}}{6}
  3. n(n+1)(2n+7)6\dfrac{{n(n + 1)(2n + 7)}}{6}
  4. n(n+1)(2n+9)6\dfrac{{n(n + 1)(2n + 9)}}{6}
Explanation

Solution

Hint : This is a basic question of sequences and series. First simplify the expression into standard form and then apply the appropriate formulas. The formulas involved in this question are:
the formula for the sum of the n consecutive integers, n(n+1)2\dfrac{{n(n + 1)}}{2} and the formula for the sum of the squares of n positive integers, n(n+1)(2n+1)6\dfrac{{n(n + 1)(2n + 1)}}{6}.

Complete step-by-step answer :
Let’s begin the question by writing out the expression we have i.e.,
⇒∑k=1nk(k+2)\Rightarrow \sum\limits_{k = 1}^n {k(k + 2)}
Now, let’s open the brackets for the expression as shown below
⇒∑k=1nk2+2k\Rightarrow \sum\limits_{k = 1}^n {{k^2} + 2k}
Now, separating the two parameters under the limit keeping everything else constant we get,
⇒∑k=1nk2+∑k=1n2k\Rightarrow \sum\limits_{k = 1}^n {{k^2}} + \sum\limits_{k = 1}^n {2k}
Now, taking the constant term out from the second term while keeping everything else constant, we \Rightarrow \dfrac{{n(n + 1)(2n + 1)}}{6} + \dfrac{{n(n + 1)}}{2} \times 2$$$$ \Rightarrow \sum\limits_{k = 1}^n {{k^2}} + 2\sum\limits_{k = 1}^n k
Clearly, the first part of the expression depicts the sum of the squares of n consecutive natural numbers while the second part depicts the sum of n consecutive natural numbers. We know the expression for the sum of n natural numbers as n(n+1)2\dfrac{{n(n + 1)}}{2} and sum of squares of n natural numbers as n(n+1)(2n+1)6\dfrac{{n(n + 1)(2n + 1)}}{6}. So, we get,
After replacing their values, we find n(n+1)n(n + 1)is common in both the expressions, therefore taking it out while keeping everything else constant, we get,
⇒n(n+1)[(2n+1)6+1]\Rightarrow n(n + 1)\left[ {\dfrac{{(2n + 1)}}{6} + 1} \right]
Now, solving the bracket we get,
⇒n(n+1)[2n+76]\Rightarrow n(n + 1)\left[ {\dfrac{{2n + 7}}{6}} \right]
And finally, we arrive at the final answer
⇒n(n+1)(2n+7)6\Rightarrow \dfrac{{n(n + 1)(2n + 7)}}{6}
So, the correct answer is “Option 3”.

Note : In mathematics, a series is the cumulative sum of a given sequence of terms. Typically, these terms are real or complex numbers, a sequence is an enumerated collection of objects in which repetitions are allowed and order matters. Like a set, it contains members. The number of elements is called the length of the sequence.