Sum of first n natural numbers

The sum of all badge numbers is (S_{38}=\frac{38\times39}{2}=741). When counting starts from (1), take the last number as (n).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). For \(n=29\), \(\frac{29\times30}{2}=29\times15=435\). Therefore, 435 is correct. Values such as 425 or 445 can result from an addition or multiplication error. Exam tip: Cancel the factor 2 with the even number first to calculate quickly.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{31}=\frac{31\times32}{2}=31\times16=496\). Option 512 results from incorrectly using 32 in place of 31. Exam tip: When finding the sum up to \(n\), remember that the formula includes \(n+1\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=33\), so \(\frac{33\times34}{2}=33\times17=561\). Option 544 can result from incorrectly using 32 instead of 33, but the sum runs from 1 to 33. Exam tip: take the last number as \(n\) and apply \(\frac{n(n+1)}{2}\) directly.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=34\), we get \(\frac{34\times35}{2}=17\times35=595\). Hence, 595 is the correct answer. A value such as 578 does not result from \(\frac{34\times35}{2}\). Exam tip: For the sum of consecutive natural numbers from 1 to \(n\), directly use \(\frac{n(n+1)}{2}\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=37\), we get \(\frac{37\times38}{2}=37\times19=703\). Therefore, 703 is correct. The values 693, 713, and 723 do not result from this formula. Exam tip: Write \(n(n+1)/2\) and simplify the even factor by 2 first.

The sum of the first \\(n\\) natural numbers is \\(\frac{n(n+1)}{2}\\). Here, \\(n=38\\), so \\(\frac{38\times39}{2}=19\times39=741\\). Therefore, 741 is correct. The option 760 may result from mistakenly using 40 instead of \\(n+1=39\\). Exam tip: use \\(\frac{n(n+1)}{2}\\) for the sum from 1 to \\(n\\).

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{39}=\frac{39\times40}{2}=39\times20=780\). Option 760 equals \(38\times20\), so it is not the sum of the first 39 natural numbers. Exam tip: For the sum from \(1\) to \(n\), use \(n(n+1)/2\).

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{41}=\frac{41\times42}{2}=41\times21=861\). Hence, 861 is the correct answer. A value such as 841 is less than the correct product \(41\times21\). Exam tip: Cancel 2 with the even factor first to calculate quickly.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=42\), we get \(\frac{42\times43}{2}=21\times43=903\). The value \(882\) comes from using \(42\times21\) and misses the required \(n+1=43\) factor. Exam tip: always include \(n+1\) when applying this sum formula.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=43\), so \(\frac{43\times44}{2}=43\times22=946\). A value such as 936 results from an arithmetic error in multiplication or division. Exam tip: For the sum from \(1\) to \(n\), use \(\frac{n(n+1)}{2}\).

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{44}=\frac{44\times45}{2}=22\times45=990\). Hence, 990 is correct. A value such as 968 results from not applying the multiplication and division correctly. Exam tip: in \(n(n+1)/2\), always use the number immediately after \(n\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=46\), we get \(\frac{46\times47}{2}=23\times47=1081\). Therefore, 1081 is correct. A value such as 1071 can result from an error in multiplication or subtraction. Exam tip: when \(n\) is even, divide it by 2 first to simplify the calculation.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{47}=\frac{47\times48}{2}=47\times24=1128\). Hence, option B is correct. Dividing \(48\) by \(2\) first makes the calculation simpler than multiplying \(47\times48\) directly. Exam tip: use \(n(n+1)/2\) for the sum of the first \(n\) natural numbers.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Therefore, \(1+2+\cdots+48=\frac{48\times49}{2}=24\times49=1176\). Hence, option B is correct. Squaring 48 or multiplying 48 and 49 without dividing by 2 gives an incorrect result. Exam tip: for the sum of consecutive natural numbers from 1 to \(n\), use \(\frac{n(n+1)}{2}\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Putting \(n=49\), we get \(\frac{49\times 50}{2}=49\times 25=1225\). Hence, 1225 is the correct answer. A value such as 1215 may result from an incorrect multiplication or addition. 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 \(n=52\), we get \(\frac{52\times53}{2}=26\times53=1378\). Hence, 1378 is the correct option. A value such as 1368 may result from an arithmetic error in multiplication or addition. Exam tip: when \(n\) is even, divide it by 2 first to calculate faster.

\(S_n\) includes terms from 1 to \(n\), whereas \(S_{n-1}\) includes terms from 1 to \(n-1\). On subtraction, only \(n\) remains. \(2n-1\) is the difference of consecutive square sums. Exam tip: subtract successive sums to identify the added term.

The direct answer is option C, 38. The sum of the first n natural numbers is found by S_n=n(n+1)/2. We need n(n+1)/2=741. Multiply by 2: n(n+1)=1482. Since 38×39=1482, n=38, and checking gives 38×39/2=741. Option A, 36, gives 36×37/2=666, so it is too small. Option B, 37, gives 37×38/2=703, also too small. Option C, 38, gives exactly 741, so it is correct. Option D, 39, gives 39×40/2=780, which is too large. Remember: for the first n natural numbers, multiply n by the next number and divide by 2.

\(n\) and \(n+1\) are consecutive integers. One consecutive integer is even and the other is odd, so their product is even and can be divided by 2. Exam tip: remember the even-odd pattern of consecutive integers.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=990\), so \(n(n+1)=1980\). Since \(44\times45=1980\), \(n=44\). For the closest distractor, \(n=43\) gives \(\frac{43\times44}{2}=946\), not 990. Exam tip: for such questions, look for two consecutive factors of \(2S\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Here, \(\frac{n(n+1)}{2}=1128\), so \(n(n+1)=2256\). Since \(47\times48=2256\), \(n=47\). Option 48 is not correct because \(\frac{48\times49}{2}=1176\). Exam tip: for such questions, look for two consecutive factors whose product is twice the given sum.

Direct answer: n = 49, so option C is correct. Here Sₙ means the sum 1 + 2 + 3 + ... + n. The formula is Sₙ = n(n + 1)/2. Substitute the given sum: n(n + 1)/2 = 1225, so n(n + 1) = 2450. Now test the natural neighboring value 49: 49 × 50 = 2450. Hence S₄₉ = 49 × 50/2 = 49 × 25 = 1225. Therefore option C is correct. To reject the other choices, calculate their sums: S₄₇ = 47 × 48/2 = 1128, S₄₈ = 48 × 49/2 = 1176, and S₅₀ = 50 × 51/2 = 1275. None equals 1225. The sum increases whenever n increases because a new positive natural number is added, so there cannot be another listed value with the same sum. Common warning: do not solve n(n + 1) = 1225; the factor 2 must be handled first.

Use the formula \(S_n=\frac{n(n+1)}{2}\) for the sum of the first \(n\) natural numbers. Thus, \(S_{29}=\frac{29\times30}{2}=435\). Therefore, 435 is the correct answer. A value such as 425 results from an incorrect calculation while applying the formula. Exam tip: for a sum from 1 to \(n\), use \(\frac{n(n+1)}{2}\).

The sum of natural numbers from 1 to \(n\) is \(\frac{n(n+1)}{2}\). Here, \(n=31\), so \(\frac{31\times 32}{2}=31\times 16=496\). Therefore, 496 is correct. The value 465 is the sum from 1 to 30, so it is a close but incorrect distractor. Exam tip: use the last number in the series as \(n\).