Sum of first n natural numbers

Use the formula for the sum of the first n natural numbers: \(S_n=\frac{n(n+1)}{2}\). Since \(S_n=6903\), we solve \(\frac{n(n+1)}{2}=6903\), so \(n(n+1)=13806\). The consecutive numbers 117 and 118 have product \(117\times118=13806\), hence n=117. The difference \(S_{n+5}-S_n\) is therefore the sum of the next five terms: 118 through 122.

Adding them gives \(118+119+120+121+122=600\). A quick check uses their average, 120, multiplied by 5: \(5 imes120=600\). Hence option B is correct. The important step is that subtracting the two partial sums removes the first n terms and leaves exactly the next five terms. The supplied answer and explanation are consistent.

There are \(225-151+1=75\) terms from 151 to 225. Their average is \(\frac{151+225}{2}=188\). Therefore, the sum is \(75\times188=14100\). A value such as 14050 can result from counting the terms or the last term incorrectly. Exam tip: for consecutive numbers, use \(\text{number of terms}\times\text{average}\) for a quick calculation.

Direct answer: option B, \(S_n=648\). We are given two equations: \(S_n+S_{n-1}=1260\) and \(S_n-S_{n-1}=36\). Add them term by term: the two \(S_{n-1}\) terms cancel, giving \(2S_n=1296\). Divide by 2: \(S_n=648\). Option A, 630, is half of 1260 but ignores the second equation, so it is not justified. Option B is exactly the calculated value. Option C, 666, and option D, 684, do not satisfy the pair of equations; for example, each would make \(2S_n\) equal to 1332 or 1368 rather than 1296. Subtraction could also be used, but addition is quickest because it removes \(S_{n-1}\). Memory cue: when one equation has a plus and the other has a minus, adding them isolates the first quantity.

Here, \(S_k\) denotes the sum of the first \(k\) natural numbers. In \(S_{n+1}-S_{n-2}\), the terms from 1 to \(n-2\) cancel, leaving \((n-1)+n+(n+1)=3n\). Thus, \(3n=330\), so \(n=110\). If 109 were used, the difference would be \(3\times109=327\), not 330. Exam tip: for differences of partial sums, cancel the common initial terms and write the remaining consecutive terms.

The sum of the first 180 natural numbers contains all the terms in the sum of the first 179 natural numbers, plus only 180. Hence, \(S_{180}-S_{179}=180\). Option 16290 is the total sum \(1+2+\cdots+180\), not the difference between the two sums. Exam tip: The difference between consecutive partial sums is always the next term.

Let the sum of the first n natural numbers be \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{250}-S_{200}=\frac{250\times251}{2}-\frac{200\times201}{2}=31375-20100=11275\). This difference is the sum of the numbers from 201 to 250. The nearby option 11175 is incorrect because it does not give the correct total of all 50 terms from 201 to 250. Exam tip: Rewrite such a difference directly as \(201+202+\cdots+250\).

(7626=S_{123}), so (S_{125}=\frac{125\cdot126}{2}=7875). First identify (n) and then move (2) ahead.

(S_n-S_{n-1}=n), so (n=150) and (S_{149}=\frac{149\cdot150}{2}=11175). For the previous sum, take (n-1).

The sum of the first n natural numbers is \\(S_n=\frac{n(n+1)}{2}\\). For n=125, \\(S_{125}=\frac{125\times126}{2}=125\times63=7875\\). The question asks for 40 percent of this sum. Since \\(40\%=\frac{40}{100}=\frac25\\), multiply \\(7875\\) by \\(\frac25\\): \\(7875\times\frac25=1575\times2=3150\\).

Therefore option A is correct. The percentage must be applied to the entire sum, not just to 125, and 40 percent is not the same as multiplying by 40. Converting the percentage to a fraction makes the calculation clear and avoids a decimal error. The result is an integer and agrees with the listed first option.

\(S_n\) contains the sum from 1 to \(n\). Subtracting \(S_{n-8}\) removes terms through \(n-8\), leaving \((n-7)+\cdots+n\), exactly 8 terms. \(S_{n+8}-S_n\) gives the next 8 terms instead. Exam tip: count the remaining terms.

The sum is \(\frac{n(n+1)}{2}\). On dividing by \(n\), we get \(\frac{n+1}{2}\), which is an integer only when \(n\) is odd. Being prime is not sufficient because 2 is even. Exam tip: cancel common factors first in divisibility questions.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_n-S_{n-6}\) is the sum of the last six numbers: \((n-5),(n-4),\ldots,n\). Hence, \(S_n-S_{n-6}=6n-15\). Using the given condition, \(6n-15=813\), so \(6n=828\) and \(n=138\). Therefore, option D is correct. If \(n=137\), the difference is \(807\), not 813. Exam tip: Interpret \(S_n-S_{n-k}\) as the sum of the last \(k\) terms to solve quickly.

Here, \(S_k\) denotes the sum of the first \(k\) natural numbers. In \(S_{n+4}-S_{n-1}\), the terms from \(1\) to \(n-1\) cancel, leaving \(n+(n+1)+(n+2)+(n+3)+(n+4)=5n+10\). Thus, \(5n+10=485\), so \(5n=475\) and \(n=95\). If \(n=96\), the difference would be \(490\), not \(485\). Exam tip: the sum of five consecutive terms equals \(5\) times their middle term, i.e. \(5(n+2)\).

The sum is \(S_n=\frac{n(n+1)}{2}\). Dividing by n gives \(\frac{n+1}{2}\), which is an integer only when n is odd. Even n leaves a half-integer. Exam tip: test divisibility by cancelling n first.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Here, \(\frac{n(n+1)}{2}=2701\) gives \(n=73\), since \(S_{73}=\frac{73\times74}{2}=2701\). Therefore, \(S_{n+7}-S_n=S_{80}-S_{73}=74+75+\cdots+80\). This is the sum of 7 consecutive numbers: \(\frac{7(74+80)}{2}=539\). Hence, 539 is correct. A value such as 535 can result from using an incorrect average or number of terms. Exam tip: \(S_{n+k}-S_n\) represents the sum of terms from \(n+1\) to \(n+k\).

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{84}=\frac{84\times85}{2}=3570\) and \(S_{56}=\frac{56\times57}{2}=1596\). Therefore, \(3570:1596=85:38\), in simplest form. \(84:56=3:2\) is only the ratio of the number of terms, not the ratio of their sums. Exam tip: For ratios of such sums, apply \(\frac{n(n+1)}{2}\) to both values of \(n\).

To find the sum from 101 to 160, subtract the sum from 1 to 100 from the sum from 1 to 160. Thus, \(S_{160}-S_{100}=\frac{160\times161}{2}-\frac{100\times101}{2}=12880-5050=7830\). Hence, 7830 is the correct answer. An option such as 7780 can result from a small subtraction or multiplication error. Exam tip: for a range of consecutive natural numbers, use \(S_n=\frac{n(n+1)}{2}\).

(16290=S_{180}), so (S_{172}=\frac{172\cdot173}{2}=14878). Identify (n) first and then reduce it by (8).

In \(S_{n+2}-S_{n-3}\), all terms up to \((n-3)\) cancel. The five remaining terms are \((n-2)+(n-1)+n+(n+1)+(n+2)=5n\). Hence, \(5n=625\), giving \(n=125\). If \(n=126\), the difference would be \(630\), so it is not correct. Exam tip: In differences of partial sums, write and add the uncancelled terms between the two indices.

\(S_n\) has the extra final term \(n\) beyond \(S_{n-1}\), so their difference is \(n\). Thus A is correct; B gives the previous term. Exam tip: consecutive sums differ by the newly added term.

The sum of the first 90 natural numbers is \(S_{90}=\frac{90\times91}{2}=4095\). Hence, \(S_n=5775+4095=9870\). Using \(\frac{n(n+1)}{2}=9870\), we get \(n(n+1)=19740=140\times141\); therefore, \(n=140\). For instance, if \(n=135\), the sum is \(\frac{135\times136}{2}=9180\), not the required \(9870\). Exam tip: In such questions, first add the known \(S_k\) to find \(S_n\).

There are \(210-151+1=60\) terms from 151 to 210. Their average is \(\frac{151+210}{2}=180.5\). Hence, the sum is \(60\times180.5=10830\). An option such as 10730 can result from an error in counting the terms or finding the average. Exam tip: when both endpoints are included, always add \(+1\) while counting the terms.

The sum of the first \(r\) natural numbers is \(S_r=\frac{r(r+1)}{2}\). Given \(S_{2n}=4095\), we have \(\frac{2n(2n+1)}{2}=4095\). Since \(4095=\frac{90\times91}{2}=S_{90}\), \(2n=90\), so \(n=45\). If \(n=46\), then the index would be 92, whose sum is not 4095. Exam tip: first identify the index corresponding to the given sum \(S_r\).

Here, \(S_n-S_{n-4}\) represents the sum of the last four natural numbers from \((n-3)\) to \(n\). Thus, \((n-3)+(n-2)+(n-1)+n=4n-6\). Hence, \(4n-6=454\), so \(4n=460\) and \(n=115\). If \(n=114\), the sum is \(450\), so it is the closest option but not correct. Exam tip: In \(S_n-S_{n-k}\), add the last \(k\) terms.

The sum of the first n natural numbers is
\(S_n=\frac{n(n+1)}{2}\). Thus,
\(S_{225}=\frac{225\times226}{2}=25425\) and
\(S_{175}=\frac{175\times176}{2}=15400\). Therefore, the difference is
\(25425-15400=10025\). It is also the sum of the integers from 176 to 225. Exam tip: the difference of two such sums equals the sum of the remaining consecutive terms.