What is the 99^{th} term of the series 2 + 7 | vedclass

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(A) Let the given series be $S = 2 + 7 + 14 + 23 + 34 + \dots + a_n$. The differences between consecutive terms are $7-2=5$,$14-7=7$,$23-14=9$,$34-23=

Vedclass · Accepted Answer

$9998$

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(A) Let the given series be $S = 2 + 7 + 14 + 23 + 34 + \dots + a_n$. The differences between consecutive terms are $7-2=5$,$14-7=7$,$23-14=9$,$34-23=11$,which form an arithmetic progression with a common difference of $2$. Let the $n^{th}$ term be $a_n = An^2 + Bn + C$. For $n=1$,$A+B+C = 2$. For $n=2$,$4A+2B+C = 7$. For $n=3$,$9A+3B+C = 14$. Subtracting the first from the second: $3A+B = 5$. Subtracting the second from the third: $5A+B = 7$. Subtracting these results: $2A = 2$,so $A = 1$. Substituting $A=1$ into $3A+B=5$,we get $B=2$. Substituting $A=1, B=2$ into $A+B+C=2$,we get $1+2+C=2$,so $C=-1$. Thus,the general term is $a_n = n^2 + 2n - 1$. For the $99^{th}$ term,$n=99$: $a_{99} = (99)^2 + 2(99) - 1 = 9801 + 198 - 1 = 9998$.

What is the $99^{th}$ term of the series $2 + 7 + 14 + 23 + 34 + \dots$?

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