Sum of first n natural numbers

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=990\), 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.

(\frac{49\cdot50}{2}=1225), so (n=49). Substituting options in the formula gives the answer quickly.

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 \(n\).

By the formula, (S_{33}=\frac{33\cdot34}{2}=561). If the list starts from (1), use the natural-number sum directly.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=37\), so the sum is \(\frac{37\times38}{2}=37\times19=703\). A value such as 693 can result from an arithmetic error while multiplying or dividing. Exam tip: for consecutive numbers from 1 to \(n\), apply \(\frac{n(n+1)}{2}\) directly.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Therefore, the sum from \(1\) to \(42\) is \(\frac{42\times43}{2}=21\times43=903\). A value such as \(893\) can result from an error in multiplication or division. Exam tip: for consecutive natural numbers starting from \(1\), directly use \(\frac{n(n+1)}{2}\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here \(n=46\), so \(\frac{46\times47}{2}=23\times47=1081\). Therefore, the correct answer is 1081. A value such as 1071 can result from an error in multiplication or halving. Exam tip: take the last number as \(n\) and apply the formula directly.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{54}=\frac{54\times55}{2}=27\times55=1485\). Hence, 1485 is the correct answer. A value such as 1475 may result from a small multiplication or addition error. Exam tip: for the sum of consecutive numbers from 1 to \(n\), use \(\frac{n(n+1)}{2}\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=58\), so \(\frac{58\times59}{2}=29\times59=1711\). Therefore, 1711 is correct. A value such as 1701 can result from an arithmetic error. 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, \(S_{65}=\frac{65\times66}{2}=65\times33=2145\). Hence, 2145 is correct. A common mistake is to calculate \(65\times66\) but forget to divide by 2. 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 \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{70}=\frac{70\times71}{2}=35\times71=2485\). Hence, option C is correct. \(2475\) may result from an arithmetic error while multiplying or dividing. Exam tip: When \(n\) is even, divide \(n\) by 2 first to simplify the calculation.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Therefore, \(1+2+\cdots+80=\frac{80\times81}{2}=40\times81=3240\). Hence, option C is correct. A value such as 3230 can result from an arithmetic error while halving or multiplying. Exam tip: For the sum of consecutive natural numbers, use \(\frac{n(n+1)}{2}\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=90\), we get \(\frac{90\times91}{2}=45\times91=4095\). Therefore, \(4095\) is the correct answer. A value such as \(4085\) may result from a small multiplication or division error. Exam tip: Cancel the even factor with 2 before multiplying \(n(n+1)\).

\(S_n\) contains the numbers from 1 to \(n\). To obtain \(S_{n+1}\), only the next natural number, \(n+1\), is added; hence A is correct. Exam tip: list the added term to verify a recurrence.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=2145\), so \(n(n+1)=4290\). Since \(65\times66=4290\), \(n=65\). For \(n=64\), the sum is \(\frac{64\times65}{2}=2080\), not 2145. Exam tip: Set the product of two consecutive numbers equal to \(2S\) to find \(n\) quickly.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Pairing the sequence gives \(2S_n=n(n+1)\). \(\frac{n(n-1)}{2}\) is the sum up to \(n-1\). Exam tip: always check for the \(n+1\) factor.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Substituting \(n=90\) gives \(\frac{90\times91}{2}=45\times91=4095\), so the correct value is \(90\). For \(n=89\), the sum is \(4005\), making it a close but incorrect option. In exams, verify the value by substituting it into the sum formula.

The sum of the first 54 natural numbers is found using \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{54}=\frac{54\times55}{2}=27\times55=1485\). Therefore, 1485 is correct. A value such as 1475 can result from an error while multiplying or dividing by 2. Exam tip: Substitute the last natural number for \(n\) in the formula.