Sum of first n natural numbers

When the sum of the first 62 natural numbers is subtracted from the sum of the first 67 natural numbers, only the terms from 63 to 67 remain. Thus, \(63+64+65+66+67=325\). Hence, option B is correct. A value of 315 would result from an incorrect addition of these terms. Exam tip: the difference of two consecutive partial sums equals the sum of the intervening terms.

Given \(\frac{n(n+1)}{2}=861\), we get \(n(n+1)=1722\). Since \(41\times42=1722\), \(n=41\). For the close distractor \(40\), the sum is \(\frac{40\times41}{2}=820\), not 861. Exam tip: look for two consecutive numbers whose product equals twice the given sum.

To find the sum from 26 to 50, subtract the sum of natural numbers up to 25 from the sum up to 50. Thus, \\(S_{50}-S_{25}=\frac{50\times51}{2}-\frac{25\times26}{2}=1275-325=950\\). Therefore, the correct answer is 950. An option such as 975 can result from using an incorrect number of terms or lower limit. Exam tip: For a range sum, always subtract the sum up to one less than the lower limit.

Here, \(S_n-S_{n-4}\) represents the sum of the four natural numbers from the \((n-3)\)th term to the \(n\)th term: \(S_n-S_{n-4}=(n-3)+(n-2)+(n-1)+n=4n-6\). Hence, \(4n-6=250\), so \(4n=256\) and \(n=64\). Therefore, 64 is the correct option. For instance, if \(n=63\), the sum is \(60+61+62+63=246\), not 250. Exam tip: \(S_n-S_{n-k}\) always gives the sum of the last \(k\) terms.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{60}=\frac{60\times61}{2}=1830\) and \(S_{40}=\frac{40\times41}{2}=820\). Therefore, the difference is \(1830-820=1010\). It is also the sum of the numbers from \(41\) to \(60\); \(1000\) does not give the correct difference. Exam tip: the difference of two such sums can be found by adding the remaining consecutive terms directly.

(1275=S_{50}), so (S_{53}=\frac{53\cdot54}{2}=1431). The correct option should be (C).

The sum is \(S_n=\frac{n(n+1)}{2}\). It is odd only when n leaves remainder 1 or 2 on division by 4. Merely knowing that n is odd or even is not enough. Exam tip: check \(n\bmod 4\).

There are \(75-51+1=25\) terms from 51 to 75, inclusive. This is an arithmetic sequence with first term 51 and last term 75. Hence, the sum is \(\frac{25}{2}(51+75)=\frac{25}{2}\times126=1575\). Therefore, option A is correct. An answer such as 1600 can result from an error in counting the terms or finding the average. Exam tip: always add \(+1\) when counting terms from one inclusive endpoint to another.

(S_n-S_{n-1}=n), so (n=57) and (S_{57}=\frac{57\cdot58}{2}=1653). Identify (n) from the difference.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{25}=\frac{25\times26}{2}=325\) and \(S_{10}=\frac{10\times11}{2}=55\). Therefore, the ratio is \(325:55=65:11\). Option 55:13 is incorrect because it is not obtained by simplifying 325:55 correctly. Exam tip: always reduce a ratio by dividing both terms by their greatest common factor.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. From \(\frac{n(n+1)}{2}=2485\), we get \(n=70\), since \(\frac{70\times71}{2}=2485\). Therefore, \(S_n-S_{n-5}=S_{70}-S_{65}\), which is the sum of the last five numbers: \(66,67,68,69,70\). Hence, \(66+67+68+69+70=340\). Choosing 350 would miss the correct first term, 66, among these five terms. Exam tip: \(S_n-S_{n-r}\) always represents the sum of the last \(r\) terms.

The sum of the first 80 natural numbers is \(\frac{80\times81}{2}=3240\), and the sum of the first 50 natural numbers is \(\frac{50\times51}{2}=1275\). Therefore, the required difference is \(3240-1275=1965\). Equivalently, it is the sum of the integers from 51 to 80. Exam tip: use \(n(n+1)/2\) carefully for both sums to avoid a close error such as 1975.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=595\), we get \(n=34\). Therefore, \(S_{n+6}-S_n=S_{40}-S_{34}=35+36+37+38+39+40=225\). Although 230 may seem close, it does not equal the sum of the next six terms. In exams, first find n and then add only the newly included terms.

For \(n=15\), the index \(2n\) becomes 30. Hence, \(S_{2n}=S_{30}=\frac{30(30+1)}{2}=\frac{30\times31}{2}=465\). Option 435 equals \(S_{29}\), so it is not correct. Exam tip: evaluate the index \(2n\) before substituting into the formula.

\(S_{2n}=\frac{2n(2n+1)}{2}=n(2n+1)\), whereas \(2S_n=n(n+1)\). Their difference is \(n^2\), so \(S_{2n}=2S_n+n^2\). Option D ignores this extra \(n^2\) term. Exam tip: substitute \(S_n=\frac{n(n+1)}{2}\) before comparing such expressions.

Adding the two equations gives (2S_n=420), so (S_n=210). When two equations are given, addition is quick.

The sum of the first n natural numbers is \(1+2+\cdots+n=\frac{n(n+1)}{2}\), half the product of consecutive integers n and \(n+1\). It is not always even: for n=2, the sum is 3. Exam tip: check the \(n+1\) factor carefully.

The sum of the first 36 natural numbers is \(S=\frac{36\times 37}{2}=666\). Therefore, \(\frac{1}{3}\) of this sum is \(666\div 3=222\). Although 224 is close, it is not obtained by dividing 666 by 3. Exam tip: first use \(\frac{n(n+1)}{2}\) to find the sum, then apply the given fraction.

(S_{56}-S_{30}=1596-465=1131), so (1110) does not match the listed options. Always check options carefully.

There are 20 numbers from 101 to 120, inclusive. Therefore, using the sum of an arithmetic progression,
\(S=\frac{20}{2}(101+120)=10\times221=2210\). Hence, option B is correct. The value 2190 is the sum from 100 to 119, so it does not match the given range. Exam tip: Count terms in an inclusive range using \(\text{last term}-\text{first term}+1\).