Sum of first n natural numbers

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{135}=\frac{135\times136}{2}=9180\) and \(S_{90}=\frac{90\times91}{2}=4095\). Therefore, the difference is \(9180-4095=5085\). Equivalently, it is the sum of the numbers from 91 to 135. Exam tip: write \(S_{135}-S_{90}\) first, then apply the sum formula.

First (n=126), and (S_m=1953) gives (m=62), so (n-m=64). The difference of consecutive sums gives the index.

(S_{115}+116+117+118+119+120=S_{120}). When consecutive new terms are added, the last term becomes the new index.

(S_{125}=\frac{125\times126}{2}=7875), so (n=125). For large values, using (2S_n) makes identification easier.

(S_{29}=435), (S_{58}=1711), and (S_{87}=3828), so the total is (5974). Add the three different indices carefully.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Hence, \(\frac{n(n+1)}{2}=10878\), so \(n(n+1)=21756\). Since \(147\times148=21756\), \(n=147\). Therefore, \(n-25=147-25=122\). Thus, option C is correct. A nearby choice such as 120 would give \(n=145\), whose sum is not 10878. Exam tip: after using \(n(n+1)/2\), look for two consecutive integers whose product equals twice the given sum.

The difference is ((n-1)+n+(n+1)+(n+2)=4n+2), and (4n+2=286) gives (n=71). Form the sum of the four middle terms.

The difference is (241+242+\cdots+250=2455). Even with large indices, only the terms in between need to be added.

(S_{135}=\frac{135\times136}{2}=9180), so (n=135). Match (2S_n) with a consecutive product.

(S_{72}-S_{36}=2628-666=1962), and (2\times1962=3924). First find the difference and then multiply.

(S_{121}=7381), so (S_{123}=7381+122+123=7626). To move two steps ahead, add the next two terms.

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 sum is (S_{125}-S_{60}=7875-1830=6045). For a middle range, subtract the sum before the starting point.

(S_{150}-S_{140}=141+142+\cdots+150=1455). The difference of nearby sums is the sum of the terms in between.

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 difference is (S_{120}-S_{60}=7260-1830=5430). In comparison questions, find both sums separately.

(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.