Sum of first n natural numbers

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=2211\), so \(n(n+1)=4422\). Since \(66\times67=4422\), \(n=66\). If \(n=67\), the sum would be \(\frac{67\times68}{2}=2278\), not 2211. Exam tip: multiply the given sum by \(2\) and look for two consecutive factors.

The sum \(1+2+3+\cdots+n\) is given by \(\frac{n(n+1)}{2}\). The expression \(\frac{n(n+1)(2n+1)}{6}\) is for the sum of squares. Exam tip: identify the sequence before selecting a formula.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{30}=\frac{30\times31}{2}=465\) and \(S_{20}=\frac{20\times21}{2}=210\). Therefore, \(S_{30}:S_{20}=465:210=31:14\). In 31:15, the second term does not match the simplified ratio. Exam tip: always divide both terms of a ratio by their greatest common factor to obtain the simplest form.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=3240\), we get \(n=80\). Hence, \(S_{n+1}-S_n=S_{81}-S_{80}=81\), because the next sum includes only the next natural number. \(80\) is a close distractor, but it is the last number included in \(S_{80}\); the added number is \(81\). Exam tip: the difference of consecutive partial sums is always the next term.

There are \(10\) numbers from 81 to 90. Their average is \(\frac{81+90}{2}=85.5\), so the sum is \(10\times 85.5=855\). Hence, option B is correct. A value such as \(845\) can result from an error in counting the terms or finding the average. Exam tip: for consecutive numbers, use \(\text{number of terms}\times\text{average of the first and last terms}\).

\(\frac{73\cdot74}{2}=2701\), so \(n=73\). Verify the options by substituting them in the formula.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). For \(n=40\), the sum is \(\frac{40\times41}{2}=820\). One-fourth of this is \(\frac{820}{4}=205\), so option B is correct. A value such as 195 results from an error in finding the sum or dividing it. Exam tip: first calculate the total using \(\frac{n(n+1)}{2}\), then take the required fraction.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{75}=\frac{75\times76}{2}=2850\) and \(S_{50}=\frac{50\times51}{2}=1275\). Therefore, the difference is \(2850-1275=1575\). Although 1585 is close, it does not result from the correct subtraction. Exam tip: this difference can also be viewed as the sum of the numbers from \(51\) to \(75\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=4950\), so \(n(n+1)=9900\). Since \(99\times100=9900\), \(n=99\). For \(n=100\), the sum would be \(5050\), not 4950. Exam tip: Multiply the given sum by 2 and look for two consecutive numbers whose product equals the result.

The sum is (S_{100}-S_{90}=5050-4095=955). The sum of the last (10) numbers is quickly found by subtraction.

(3160=S_{79}), so (S_{79}-S_{74}=75+76+77+78+79=385), not (390). Always match options with the final calculation.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=47\), we get \(\frac{47\times48}{2}=47\times24=1128\). Therefore, option B is correct. A value such as \(1104\) can result from using an incorrect number of terms or multiplication. Exam tip: always write both \(n\) and \(n+1\) in the formula.

(S_{33}=\frac{33\times34}{2}=561), so (n=33). Checking options in the formula is fast.

The sum is (S_{29}-S_{11}=435-66=369). When the sequence does not start from (1), subtract the previous terms' sum.

Total questions are (S_{38}=\frac{38\times39}{2}=741). A daily increase of (1) forms the sum of natural numbers.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{22}=\frac{22\times23}{2}=253\) and \(S_{17}=\frac{17\times18}{2}=153\). Therefore, \(S_{22}+S_{17}=253+153=406\). Although 396 is close, it is not obtained by adding the two correct sums. Exam tip: whenever \(S_n\) appears, use \(\frac{n(n+1)}{2}\) directly.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{41}=\frac{41\times42}{2}=861\) and \(S_{23}=\frac{23\times24}{2}=276\). Therefore, \(S_{41}-S_{23}=861-276=585\). This is also the sum of the natural numbers from 24 to 41. Option 575 is only a nearby value, not the correct difference. Exam tip: You may also directly find the sum from 24 to 41 in such questions.

(S_{43}=\frac{43\times44}{2}=946), so (n=43). Match (2S_n) with the product of two consecutive numbers.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{35}=\frac{35\times36}{2}=630\) and \(S_{19}=\frac{19\times20}{2}=190\). Therefore, \(S_{35}-S_{19}=630-190=440\). It is also the sum of the natural numbers from 20 to 35. Exam tip: \(S_b-S_a\) represents the sum from \(a+1\) to \(b\).

Total seats are (S_{26}=\frac{26\times27}{2}=351). If seats equal the row number, use (S_n).