Sum of first n natural numbers

(6903=S_{117}), so the difference is (118+119+120+121+122=600). Take the sum of the next five terms.

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.

Adding the two equations gives (2S_n=1296), so (S_n=648). Adding equations gives (S_n) quickly.

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).

(S_{125}=7875) and (40%=\frac{2}{5}), so the value is (3150). Convert percentage into a fraction.

(\frac{225\cdot226}{2}=25425), so (n=225). Check the answer by substituting options in the formula.

\(S_n\) 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.

\(\frac{122\cdot123}{2}=7503\), so \(n=122\). For a large sum, check options in the formula.

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 \(n\).

Adding the two equations gives (2S_n=5700), so (S_n=2850). Adding equations is a quick method.

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