Sum of first n natural numbers

Total saving is (S_{31}=\frac{31\times32}{2}=496) rupees. A daily increase of (1) forms the sum of natural numbers.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(S_{18}=\frac{18\times19}{2}=171\). Three times this sum is \(3\times171=513\), so 513 is correct. A nearby value such as 503 results from not applying the sum formula correctly. Exam tip: first find the sum using \(\frac{n(n+1)}{2}\), then apply the required multiplication.

The sum of the first 26 natural numbers is
\(S_{26}=\frac{26\times27}{2}=351\). Therefore, subtracting 26 gives
\(351-26=325\). This is also the sum of the first 25 natural numbers. For example, 315 is not the correct sum of the first 25 natural numbers. Exam tip: remember that the sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\).

These are 10 consecutive natural numbers from 16 to 25. Their average is \(\frac{16+25}{2}=20.5\), so the sum is \(10\times20.5=205\). A value such as 195 can result from counting the number of terms or the average incorrectly. Exam tip: for consecutive terms, first count them using \((25-16+1)\).

The sum of the first n natural numbers is
\(S_n=1+2+\cdots+n\), whereas
\(S_{n-1}=1+2+\cdots+(n-1)\). Therefore,
\(S_n-S_{n-1}=n\). Since the given difference is 37,
\(n=37\). For n = 36 or 38, the difference would be 36 or 38 respectively. Exam tip: the difference between two consecutive partial sums is the newly added term.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{55}=\frac{55\times56}{2}=55\times28=1540\). Hence, 1540 is the correct option. The value \(1485\) is the sum up to 54, since \(\frac{54\times55}{2}=1485\). Exam tip: divide the even factor by 2 first to calculate quickly.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. Therefore, \(S_{25}=S_{23}+24+25=276+49=325\). Hence, the correct option is \(325\). A value such as \(300\) can result from the mistake of adding only one new term. Exam tip: When moving from \(S_n\) to \(S_{n+k}\), add every intervening new term.

If \(S_n\) denotes the sum of the first \(n\) natural numbers, then adding the next two numbers to \(S_{29}\) gives \(S_{29}+30+31=S_{31}\). \(S_{30}\) includes only up to 30, so it is not correct. Exam tip: when consecutive next terms are added to a partial sum, increase the subscript accordingly.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{20}=\frac{20\times21}{2}=210\) and \(S_{10}=\frac{10\times11}{2}=55\). Hence, \(S_{20}:S_{10}=210:55=42:11\). The ratio \(21:5\) would result from incorrectly taking \(S_{10}\) as 50. Exam tip: simplify a ratio at the end by dividing both terms by their greatest common factor.

Total beads are (S_{19}=\frac{19\times20}{2}=190). Triangular rows often give the sum of natural numbers.

The difference of two consecutive sums is the new term, so (S_{41}-S_{40}=41). You can answer without finding the full sums.

The sum of the first r natural numbers is \\(\frac{r(r+1)}{2}\\). Thus, \\(\frac{r(r+1)}{2}=153\\), so \\(r(r+1)=306=17\times18\\) and \\(r=17\\). Therefore, \\(r+2=17+2=19\\). Option 18 is only the value of \\(r+1\\), not \\(r+2\\). Exam tip: when a sum is given, first use \\(\frac{n(n+1)}{2}\\) to find n.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{46}=\frac{46\times47}{2}=23\times47=1081\). Hence, 1081 is correct. An option such as 1071 can result from an arithmetic error in multiplication or addition. Exam tip: for the sum of consecutive natural numbers, directly use \(\frac{n(n+1)}{2}\).

Use the formula S_n = n(n + 1)/2 for the sum of the first n natural numbers. Thus S₁₂ = 12×13/2 = 78, S₁₃ = 13×14/2 = 91, and S₁₄ = 14×15/2 = 105. Their total is 78 + 91 + 105 = 274, so option A is correct. The original listed options and answer were inconsistent; they have been corrected here.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{a(a+1)}{2}=351\), we get \(a=26\), since \(\frac{26\times27}{2}=351\). Similarly, \(\frac{b(b+1)}{2}=465\) gives \(b=30\), since \(\frac{30\times31}{2}=465\). Therefore, \(b-a=30-26=4\). Option 3 is incorrect because it is one less than the actual difference between the two indices. Exam tip: equate each given sum to \(\frac{n(n+1)}{2}\) to find its index.

This becomes (S_{31}), and (S_{31}=496). When consecutive terms are added, the last term becomes the new (n).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=70\), we get \(\frac{70\times71}{2}=35\times71=2485\). Therefore, 2485 is correct. A value such as 2475 results from an arithmetic error in multiplication or addition. Exam tip: cancel 2 with an even factor in \(n(n+1)\) before multiplying.

(S_{24}=\frac{24\times25}{2}=300), so (n=24). After finding the answer, verify it in the formula.

(S_{37}=703) and (S_{38}=741), so the sum is (1444). Notice the one-term difference in nearby sums.

The governing idea is cancellation in sums of consecutive natural numbers. Subtracting S_60 from S_64 removes the terms 1 through 60 and leaves only 61 + 62 + 63 + 64. Pairing them gives (61 + 64) + (62 + 63) = 125 + 125 = 250. The formula confirms this: S_64 - S_60 = [64×65/2] - [60×61/2] = 2080 - 1830 = 250. Therefore option C is correct. The distractors may result from omitting one of the four remaining terms or making an arithmetic error in the standard sum formula.