Sum of first n natural numbers

(300=S_{24}), so (S_{27}=\frac{27\cdot28}{2}=378). Identify (n) first and then move (3) ahead.

(S_n-S_{n-3}=(n-2)+(n-1)+n=3n-3). From (3n-3=114), (n=39), so no listed option is correct.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Pairing terms gives 1+n, 2+(n-1), and so on, each totaling \(n+1\). \(\frac{n(n-1)}{2}\) is the sum only up to \(n-1\). Exam tip: check the last term before selecting a formula.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Multiplying the average, \(\frac{1+n}{2}\), by the number of terms \(n\) gives this result. \(n^2\) is the sum of the first \(n\) odd numbers. In exams, check the division by 2 carefully.

Subtracting the sum of the first 90 natural numbers from the sum of the first 100 leaves only the numbers from 91 to 100. Thus, the sum is \(91+92+\cdots+100=\frac{(91+100)\times10}{2}=955\). Therefore, option B is correct. 945 is a close distractor, but it does not use the correct average, \(95.5\), of the numbers from 91 to 100. Exam tip: In such differences, cancel the common initial terms and write the remaining range directly.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Here, \(1176=\frac{48\times49}{2}\), so \(n=48\). Therefore, \(S_{n-1}=S_{47}=\frac{47\times48}{2}=1128\). The value 1176 is \(S_{48}\), not the sum up to the previous term. Exam tip: identify the given sum as half the product of two consecutive numbers first.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{15}=\frac{15\times16}{2}=120\) and \(S_{10}=\frac{10\times11}{2}=55\). Therefore, \(120:55=24:11\). Although \(11:5\) is close, it is not the simplified form of \(120:55\). Exam tip: always divide both terms of a ratio by their greatest common factor to simplify it.

(231=S_{21}), so (S_{25}=\frac{25\cdot26}{2}=325). Identify the given (n) and move (4) ahead.

The sum is (S_{55}-S_{45}=1540-1035=505). You can also use the average of (10) consecutive terms.

The options should be checked carefully because (S_n-S_{20}=250) does not match the listed values. No listed option gives exactly (250).

The sum of the first n natural numbers is
\(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{28}=\frac{28\times29}{2}=406\) and \(S_{14}=\frac{14\times15}{2}=105\). Therefore, the difference is \(406-105=301\). An answer such as 291 usually results from an arithmetic error while subtracting the sum up to 14. Exam tip: for a difference of sums, calculate both sums using the formula and subtract the smaller one from the larger one.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=1540\), so \(n(n+1)=3080\). Since \(55\times56=3080\), \(n=55\). For example, if \(n=54\), the sum is \(\frac{54\times55}{2}=1485\), not 1540. Exam tip: find two consecutive factors of \(2S\) in such questions.

There are 10 numbers from 71 to 80, inclusive. Using the sum formula for an arithmetic sequence:

\(\frac{10}{2}(71+80)=5\times151=755\).

Therefore, the correct answer is 755. A choice such as 745 may result from an error in counting the number of terms or finding the average. Exam tip: the number of terms from one endpoint to another is \(\text{last term}-\text{first term}+1\).

The sum of the first n natural numbers is \\(\frac{n(n+1)}{2}\\). Thus, the sum of the first 20 natural numbers is \\(\frac{20\times21}{2}=210\\), and the sum of the first 5 natural numbers is \\(\frac{5\times6}{2}=15\\). Therefore, the required total is \\(210+15=225\\). The value 220 does not equal the sum of the first 20 natural numbers and may result from an incorrect calculation. Exam tip: calculate each sum separately before adding them.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=2850\), we get \(n=75\). Therefore, \(S_{n-5}=S_{70}=\frac{70\times71}{2}=2485\). The value 2556 is \(S_{71}\), so it is not correct here. Exam tip: first find \(n\) from the given sum, then calculate the sum for the required index.

(S_n-S_{n-1}=n), so (n=68) and (S_{68}=\frac{68\cdot69}{2}=2346). Identify (n) from the difference.

When the sum is extended to \(n+1\) terms, only the new number \(n+1\) is added to the first \(n\) terms. Hence, \(S_{n+1}-S_n=n+1\). Option B gives the previous last term, not the added term. Exam tip: identify the newly included term.

If \(S_n\) denotes the sum of the first \(n\) natural numbers, then \(S_{36}=1+2+\cdots+36\). Adding the next two consecutive terms, 37 and 38, gives \(1+2+\cdots+36+37+38=S_{38}\). \(S_{37}\) includes terms only up to 37, so it is not correct. Exam tip: each next term added to a partial sum increases its subscript by 1.

Using \(S_n=\frac{n(n+1)}{2}\), one of \(n\) and \(n+1\) is even. After division by 2, the product is odd only when \(n\) leaves remainder 1 or 2 on division by 4. Thus, merely calling \(n\) odd or even is insufficient. Exam tip: test residues 0, 1, 2 and 3 modulo 4.

When the sum of the first 85 natural numbers is subtracted from the sum of the first 90 natural numbers, only the numbers from 86 to 90 remain. Thus, the difference is 86 + 87 + 88 + 89 + 90 = 440. Therefore, option C is correct. The value 450 would result from incorrectly treating each of the five terms as 90. Exam tip: In differences of such sums, cancel the common initial terms and add only the remaining consecutive terms.