Sum of first n natural numbers

Correct answer: C. (561)

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

Correct answer: C. 703

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.

Correct answer: C. 903

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 \(\frac{n(n+1)}{2}\).

Correct answer: C. 1081

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.

Correct answer: C. 1485

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 \(\frac{n(n+1)}{2}\).

Correct answer: C. 1711

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.

Correct answer: C. 2145

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 \(\frac{n(n+1)}{2}\).

Correct answer: C. 2485

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.

Correct answer: C. 3240

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 \(\frac{n(n+1)}{2}\).

Correct answer: C. 4095

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 \(n(n+1)\).

Correct answer: A. \(S_{n+1}=S_n+(n+1)\)

\(S_n\) 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.

Correct answer: C. 65

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.

Correct answer: A. \(\frac{n(n+1)}{2}\)

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.

Correct answer: C. \(90\)

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.

Correct answer: C. 1485

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.

Correct answer: C. (1711)

(S_{58}=\frac{58\cdot59}{2}=1711). For a list starting from (1), use the natural-number sum formula.

Correct answer: C. 2145

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=65\), so \(\frac{65\times66}{2}=65\times33=2145\). Therefore, the correct answer is 2145. A value such as 2112 can result from using 65 or 66 incorrectly in the calculation. Exam tip: for a sum ending at \(n\), use \(\frac{n(n+1)}{2}\).

Correct answer: C. 2485

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=70\), so \(1+2+\cdots+70=\frac{70\times71}{2}=35\times71=2485\). Therefore, option C is correct. A value such as 2475 can result from a small multiplication or division error. Exam tip: divide the even factor by 2 first to calculate quickly.

Correct answer: C. 3240

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Here, \(n=80\), so \(S_{80}=\frac{80\times81}{2}=40\times81=3240\). Therefore, 3240 is the correct option. 3200 can result from the common error of using \(80\times80/2\); the formula must include \(n+1\). Exam tip: Substitute the last number for n and use \(n(n+1)/2\).

Correct answer: C. 2628

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Hence, \(S_{72}=\frac{72\times73}{2}=36\times73=2628\). Therefore, 2628 is correct. A value such as 2618 can result from an arithmetic error in multiplication or addition. Exam tip: when \(n\) is even, divide \(n\) by 2 first to simplify the calculation.

Correct answer: C. 3655

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=85\), we get \(\frac{85\times86}{2}=85\times43=3655\). Therefore, option C is correct. A value such as \(3645\) results from an arithmetic error in the calculation. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum of consecutive natural numbers starting from 1.

Correct answer: A. प्रथम n प्राकृतिक संख्याओं में सबसे बड़ी संख्या के बाद आने वाली संख्या

The first n natural numbers run from 1 to n, so n is the largest number and \(n+1\) is the next natural number. Option B represents only the largest number, n. Exam tip: identify what each factor means before using a formula.

Correct answer: A. यह सदैव सम होता है।

\(n\) and \(n+1\) are consecutive integers, so one of them must be even. Hence \(n(n+1)\) is always even, making division by 2 valid. Exam tip: consecutive integers always alternate between even and odd.

Correct answer: C. 4560

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=95\), we get \(\frac{95\times96}{2}=95\times48=4560\). Therefore, option C is correct. A value such as \(4550\) may result from an error while multiplying the middle terms. Exam tip: cancel 2 with the even factor before multiplying.

Correct answer: B. Quadratic function

The sum is \(S_n=\frac{n(n+1)}{2}\). Writing it as \(\frac{n^2+n}{2}\) shows that the highest power of n is 2, so it is a quadratic function. A linear function has highest power 1. Exam tip: expand the formula and check the highest power.