Compute the value of the sum $7\cdot 11+11\cdot 15+15\cdot 19+\cdots+95\cdot 99$

Question

Solution: My attempt: =$9^2-2^2 +13^2-2^2+17^2-2^2...97^2-2^2$
=$(9^2+13^2+17^2...97^2)-2^2( 23 )$ Focus on the first part. 81+169+289+441...Here first-order differences are 88, 120, 152... and the second-order difference is common to be 32. Therefore the series is of the form of $p(n)^2+q(n)+r$. If we plug n=0, we get the first term, if plug n=1, we get the second term, if we plug n=2, we get the third term and so on. $p*0^2+q*0+r=81$ r=81 $p*1^2*q*1+81=169$ $p+q+81=169$ $p+q=88…(1)$ $p*2^2+q*2+81=289$ $4p+2q=208…(2)$ Solving (1) and (2), we get p =16, q =72, and r =81.

Therefore the general term formula is $16n^2+72n+81$. Now, if we start adding them up, 81 will be there 23 times. Then there will be 72, 144, 216 etc, 22 times. This sum will be 72(1+2+3+...22) =18216. At the last 16n^2 will be there 22 times. This sum will be $16(1^2+2^2+3^2...22^2) = 16 ((22) (23) (45)/6) = 60720$. Grand total is $1863+18216+60720-92= 80707$.

Answer 1

We have one more way to do it easily. As we can see the sequence has the general term $T_n$=$(4n+3)(4n+7)$ and the sum of sequence will be equal to $\sum {T_n} $ from $n=1$ to $n=23$. Now, the sum will appear as : $$16\frac{n(n+1)(2n+1)}{6}+20n(n+1)+21n$$ where putting n=23 will give $80707$

It will be easier as there will be less chance for error as there will be less steps than what you have approached the question. Also what fellow users have given, it is just the same but easier coefficients to apply for.

Answer 2

\begin{align} S &= \sum_{n=0}^{n=22} (4n+7)(4n+11)\\ &= \sum_{n=0}^{n=22} 16n^2+72n+\color{red}{77}\\ &= 80707 \end{align}