Sum of first n natural numbers

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_{144}=10440) and (2S_{72}=5256), so the value is (5184). Keep the order of multiplication and subtraction clear.

Total lamps are \(6S_{45}=6\times1035=6210\). If there is a common multiplier, keep it outside the sum.

(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.

(S_{102}=5253), (S_{103}=5356), and (S_{101}=5151), so the value is (5458). Subtract carefully with nearby indices.