Sum of first n natural numbers

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Therefore, \(S_{54}=\frac{54\times55}{2}=1485\). Twice this sum is \(2\times1485=2970\). \(2916\) is \(54^2\), so it is not twice the required sum. Exam tip: when the question asks for twice the sum, the 2 cancels in \(\frac{n(n+1)}{2}\), giving \(n(n+1)\).

The sum of the last (6) terms is (35+36+37+38+39+40=225). For last terms, finding full (S_{40}) is not necessary.

To obtain the sum of the first 34 natural numbers, add the numbers after 31, namely 32, 33, and 34, to the given sum. Thus, \(S_{34}=496+32+33+34=595\). Therefore, 595 is correct. Adding only 34 would be incorrect because 32 and 33 are also new terms. Exam tip: When moving from \(S_n\) to \(S_{n+k}\), add every intervening term.

When the sum of the first 55 natural numbers is subtracted from the sum of the first 58 natural numbers, only 56, 57, and 58 remain. Thus, the difference is 56 + 57 + 58 = 171. Option 174 is 3 too large. Exam tip: to find \(S_n-S_m\), add the terms from \(m+1\) to \(n\).

(S_{46}=\frac{46\times47}{2}=1081), so (n=46). Checking with options is useful for larger sums.

Total dots are (S_{42}=\frac{42\times43}{2}=903). Having (r) dots in the (r)th row is the sum of natural numbers.

The sum of the first 28 natural numbers is \(S_{28}=\frac{28\times29}{2}=406\). Therefore, \(S_{28}+28=406+28=434\). Option 435 is \(S_{29}\), because 29—not 28—is added to 406. Exam tip: Use \(S_n=\frac{n(n+1)}{2}\), then add the separately given number carefully.

The sum of the first 65 natural numbers is \(\frac{65\times66}{2}=2145\). Therefore, the required value is \(2145-1500=645\), so option D is correct. A nearby value such as 640 is incorrect because no extra rounding is done after finding the sum and subtracting. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum of the first \(n\) natural numbers.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{p(p+1)}{2}=378\), we get \(p=27\), and from \(\frac{q(q+1)}{2}=528\), we get \(q=32\). Therefore, \(q-p=32-27=5\). Option 4 may seem close, but the corresponding indices for 378 and 528 are 27 and 32 respectively. Exam tip: write the sum as \(\frac{n(n+1)}{2}\) and identify the consecutive factors.

There are \(36-18+1=19\) terms from 18 to 36. Using the sum of an arithmetic progression, \(S=\frac{n}{2}(a+l)=\frac{19}{2}(18+36)=19\times27=513\). Hence, option A is correct. A value such as 523 results from using an incorrect number of terms or average. Exam tip: Always add \(+1\) when counting terms from one endpoint to another, inclusive.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{57}=\frac{57\times58}{2}=57\times29=1653\). Hence, 1653 is correct. A value such as 1623 can result from an error while multiplying or halving. Exam tip: remember to use \(n+1\) in the formula.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. From \(\frac{n(n+1)}{2}=666\), we get \(n=36\). Therefore, \(S_{n+2}=S_{38}=S_{36}+37+38=666+75=741\). Option 740 is one less and does not include the correct sum of the next two natural numbers. Exam tip: to find \(S_{n+2}\), add \(n+1\) and \(n+2\) to \(S_n\).

The sum of the first 72 natural numbers includes the last number, 72. Therefore, subtract 72 from 2628 to get the sum of the first 71 natural numbers: \(2628-72=2556\). Option B results from the common mistake of subtracting 71; the number removed is 72. Exam tip: to obtain \(S_{n-1}\) from \(S_n\), subtract \(n\).

(S_{24}=300) and (S_{26}=351), so the sum is (651). Read similar-looking indices carefully.

(S_{48}=\frac{48\times49}{2}=1176), so (n=48). Verify the answer by substituting it back in the formula.

The marks in the first 20 stages are the sum of natural numbers from 1 to 20: \(S_{20}=\frac{20\times21}{2}=210\). The next 5 stages contribute 0 marks, so they do not change the total. Therefore, the total score is 210. A common error is getting 200 by forgetting the \(n+1\) factor in the sum formula. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum from 1 to \(n\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=63\), we get \(\frac{63\times64}{2}=63\times32=2016\). Hence, option B is correct. An answer such as \(1996\) usually results from an error in multiplication or division. Exam tip: for \(1+2+\cdots+n\), directly use \(\frac{n(n+1)}{2}\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{44}=\frac{44\times45}{2}=990\) and \(S_{30}=\frac{30\times31}{2}=465\). Therefore, \(S_{44}-S_{30}=990-465=525\). This difference is the sum of the natural numbers from 31 to 44. A nearby option such as 535 can result from an arithmetic error in addition or subtraction. Exam tip: interpret \(S_{44}-S_{30}\) directly as the sum from 31 to 44.

\(S_{55}=\frac{55\times56}{2}=1540\), so \(n=55\). For a large sum, identify using \(2S_n\).

(1+2+\cdots+37+38+39=S_{39}). When consecutive new terms are added, the last term becomes the new (n).