Sum of first n natural numbers

The difference is ((n-4)+(n-3)+(n-2)+(n-1)+n=5n-10), so (5n-10=390) gives (n=80). Form the sum of the last five terms.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{150}=\frac{150\times151}{2}=75\times151=11325\). Hence, 11325 is correct. A value such as 11225 may result from incorrectly using \(150\times150/2\); the formula must include \(n+1\). Exam tip: write \(\frac{n(n+1)}{2}\) first, then simplify the calculation.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers, so \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{100}=5050\), \(S_{90}=4095\), and \(S_{10}=55\). Therefore, \(S_{100}-S_{90}+S_{10}=5050-4095+55=1010\). The value 955 is only \(S_{100}-S_{90}\); \(S_{10}\) must still be added. Exam tip: \(S_a-S_b\) can also be checked as the sum of integers from \(b+1\) to \(a\).

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Here, \(S_{3k}=1176\), so \(\frac{3k(3k+1)}{2}=1176\). This gives \(3k=48\), since \(\frac{48\times49}{2}=1176\). Hence, \(k=\frac{48}{3}=16\). If \(k=15\), the index would be \(45\), whose sum is \(1035\), not \(1176\). Exam tip: first find the index \(n\) from the sum, then use \(n=3k\).

These are 30 consecutive natural numbers from 71 to 100. Therefore, their sum is \(\frac{30}{2}(71+100)=15\times171=2565\). Hence, option B is correct. A value such as 2535 can result from an error in counting the number of terms or finding the average. Exam tip: for consecutive numbers, first count the terms using \((\text{last}-\text{first}+1)\).

(S_{60}=1830), so the difference is (61+62+\cdots+70=655). Find the sum of the next (10) terms separately.

(S_{99}=4950) and (S_{100}=5050), so it first exceeds (5000) at (n=100). In boundary questions, check nearby values.

The numbers of vehicles in the rows are 2, 4, 6, ..., 80, so the number in each row is twice its row number. Hence the total is \(2(1+2+3+\cdots+40)\). Since \(1+2+\cdots+40=\frac{40\times41}{2}=820\), the total is \(2\times820=1640\) vehicles. The value 1600 may result from using \(40\times40\), but that is not the correct sum of this arithmetic progression. Exam tip: identify the first term, last term and number of terms, then use \(S_n=\frac{n}{2}(a+l)\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\), so option A gives the set of triangular numbers. For example, \(n=3\) gives 6. Squares form a different set. In exams, identify the standard sum formula before choosing.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers, so \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{33}=\frac{33\times34}{2}=561\), \(S_{66}=\frac{66\times67}{2}=2211\), and \(S_{22}=\frac{22\times23}{2}=253\). Therefore, \(S_{33}+S_{66}-S_{22}=561+2211-253=2519\). An answer such as 2496 can result from calculating one of the sums or the subtraction incorrectly. Exam tip: evaluate each \(S_n\) separately before combining the results.

The sum of the first 120 natural numbers is
\(S_{120}=\frac{120\times121}{2}=7260\), and the sum of the first 80 natural numbers is
\(S_{80}=\frac{80\times81}{2}=3240\). Therefore, the difference is
\(7260-3240=4020\). It is also the sum of the numbers from 81 to 120. A value such as 4060 can result from taking an incorrect starting or ending term. Exam tip: use \(S_n=\frac{n(n+1)}{2}\) for each sum before subtracting.

First (n=84), and (S_m=990) gives (m=44), so (n-m=40). The difference of consecutive sums gives the index.

(S_{75}+76+77+78+79+80=S_{80}). When consecutive new terms are added, the last term becomes the new index.

(S_{76}=\frac{76\times77}{2}=2926), so (n=76). For large values, using (2S_n) makes identification easier.

(S_{19}=190), (S_{39}=780), and (S_{59}=1770), so the total is (2740). Add the three different indices carefully.

The sum of the first n natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=4656\), so \(n(n+1)=9312\). Since \(96\times97=9312\), \(n=96\). Therefore, \(n-10=96-10=86\). Choosing 84 would mean \(n=94\), whose sum is not 4656. Exam tip: when a sum is given, first find n using \(n(n+1)/2\), then evaluate the expression asked.

Let \\(S_k\\) denote the sum of the first \\(k\\) natural numbers. The difference \\(S_{n+1}-S_{n-1}\\) contains exactly the terms that are present in the first sum but absent from the second. Those two terms are \\(n\\) and \\(n+1\\). Therefore, \\(S_{n+1}-S_{n-1}=n+(n+1)=2n+1\\). The question gives this difference as 151, so \\(2n+1=151\\). Subtracting 1 gives \\(2n=150\\), and dividing by 2 gives \\(n=75\\). Hence option B is correct.

The same result follows from the formula \\(S_k=k(k+1)/2\\), but cancellation of the common terms is simpler and less error-prone. The first sum ends at \\(n+1\\), while the second ends at \\(n-1\\), so only two consecutive terms remain. Values 74 and 76 would make the difference 149 and 153 respectively, not 151. Thus, the supplied answer is mathematically consistent.

The difference is (191+192+\cdots+200=1955). Even with large indices, only the terms in between need to be added.

(S_{111}=\frac{111\times112}{2}=6216), so (n=111). Use (2S_n) to identify consecutive numbers.

(S_{48}-S_{24}=1176-300=876), and (3\times876=2628). First find the difference and then multiply.