Sum of first n natural numbers

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{92}=\frac{92\times93}{2}=46\times93=4278\). Hence, 4278 is correct. A value such as 4258 can result from a small multiplication or addition error. Exam tip: simplify by dividing the even number, 92, by 2 first.

(S_{99}=\frac{99\times100}{2}=4950), so (n=99). Match (2S_n) with the product of two consecutive numbers.

The direct answer is option C, \(1455\). Since \(S_{150}=1+\cdots+150\) and \(S_{140}=1+\cdots+140\), subtraction leaves \(141+142+\cdots+150\), which has 10 terms. Pair the ends: \(141+150=291\), \(142+149=291\), and so on. There are five pairs, giving \(5\times291=1455\). Option C is correct. Option A, 1435, is 20 too small. Option B, 1445, is 10 too small. Option D, 1465, is 10 too large. Formula verification: \(S_{150}-S_{140}=150\times151/2-140\times141/2=11325-9870=1455\). Exam cue: a difference of natural-number sums leaves only the terms after the smaller index.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=2080\), we get \(n=64\). Hence, \(n+9=73\), so \(S_{n+9}=S_{73}=\frac{73\times74}{2}=2701\). A value such as 2651 may result from using an incorrect index or an addition error. Exam tip: first find \(n\) from the given sum, then substitute the new index.

(S_{90}=S_{89}+90) and (S_{88}=S_{89}-89), so the value is (1). Use differences of consecutive sums.

The sum of the first n natural numbers is \\(S_n=\frac{n(n+1)}{2}\\). The question asks for how much larger the first 120-number sum is than the first 60-number sum, so we must subtract the smaller sum from the larger one. This avoids confusing the requested difference with either individual total.

For 120 numbers, \\(S_{120}=\frac{120\times121}{2}=7260\\). For 60 numbers, \\(S_{60}=\frac{60\times61}{2}=1830\\). Hence the required difference is \\(7260-1830=5430\\). Equivalently, it is the sum of the numbers from 61 through 120. Thus option C, 5430, is correct; values such as 5380 or 5480 result from an arithmetic error.

(S_{154}=11935) and (S_{155}=12090), so it first exceeds (12000) at (n=155). In boundary questions, check nearby values.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{90}=\frac{90\times91}{2}=4095\) and \(S_{45}=\frac{45\times46}{2}=1035\). Hence, \(S_{90}:S_{45}=4095:1035=91:23\). The option \(90:23\) fails to account correctly for the \(n+1\) factor. Exam tip: for such ratios, write the sum formula first and cancel common factors before calculating fully.

The difference is ((n-4)+(n-3)+(n-2)+(n-1)+n=5n-10), and (5n-10=535) gives (n=109). Form the sum of the last five terms.

Total blocks are (S_r=3160), and (S_{79}=3160), so there are (79) rows. Such figures form a triangular number.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{a(a+1)}{2}=3003\), we get \(a=77\), since \(\frac{77\times78}{2}=3003\). Similarly, \(\frac{b(b+1)}{2}=5050\) gives \(b=100\). Therefore, \(b-a=100-77=23\). Although 22 is close, it is not the difference between the correct indices. Exam tip: when a sum is given, look for consecutive integers satisfying \(n(n+1)/2\).

This is an arithmetic sequence from 86 to 140. The number of terms is \(140-86+1=55\), with first term 86 and last term 140. Therefore, the sum is \(\frac{55}{2}(86+140)=\frac{55}{2}\times226=6215\). A value such as 6185 can result from an error in counting the terms or finding the average. Exam tip: for consecutive integers, always use \(n=\text{last}-\text{first}+1\).

(S_{m+3}-S_m=(m+1)+(m+2)+(m+3)=3m+6), so (m=92). In a three-step difference, add the next three terms.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{200}=\frac{200\times201}{2}=20100\) and \(S_{160}=\frac{160\times161}{2}=12880\). Therefore, the required difference is \(20100-12880=7220\). It is also the sum of the numbers from 161 to 200, so an option such as 7120 is incorrect. Exam tip: You can verify such a difference by recognising the remaining sequence from 161 to 200.

(S_{172}=14878) and (S_{173}=15051), so it first exceeds (15000) at (n=173). In boundary questions, check the previous and next sums.

(S_{42}=903), (S_{84}=3570), and (S_{126}=8001), so the total is (12474). Write all three sums separately and add.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=5995\), we get \(n=109\), since \(\frac{109\times110}{2}=5995\). Hence, \(n-8=101\), so \(S_{n-8}=S_{101}=\frac{101\times102}{2}=5151\). The value \(5050\) is \(S_{100}\), so it is close but not correct. Exam tip: first determine n from the given sum, then calculate the sum for the required index.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). For \(n=132\), \(\frac{132\times133}{2}=66\times133=8778\). Therefore, 8778 is correct. A value such as 8712 can result from using an incorrect number in place of \(n+1\). Exam tip: Divide the even number 132 by 2 before multiplying.

(S_{81}=3321), (S_{108}=5886), and (S_{54}=1485), so the value is (7722). In mixed sums, write each (S_n) separately.

(S_{125}=7875), so (4p=125) gives no integer (p). In such questions, check divisibility of the index too.

(S_{120}=\frac{120\times121}{2}=7260), so (n=120). Match (2S_n) with the product of consecutive numbers.

The expression is ((n-1)+n+(n-7)+(n-6)=4n-14), and (4n-14=312) gives no integer (n). Convert into terms and check parity first.

If \(S_n\) denotes the sum of the first \(n\) natural numbers, the required number is \(S_{105}-S_{95}\). Thus, we must add \(96+97+\cdots+105\). There are 10 terms, with average \(\frac{96+105}{2}=100.5\), so their sum is \(10\times100.5=1005\). A distractor such as 1015 can result from incorrectly counting the terms or the last term. Exam tip: the difference of two partial sums leaves only the intervening terms.

(S_{68}=2346) and (S_{115}=6670), so (u+v=183). Identify the indices of both triangular numbers.

(S_{144}=10440), and (12.5%=\frac{1}{8}), so the value is (1305). Converting percent to a fraction makes it easier.