Let a 1= b 1=1, a n = a n -1+2 and bn=an+bn-1 for every natural number n ≥ 2. Then displaystyle∑n=115 an ⋅ bn is equal to

Question

Let a 1= b 1=1, a n = a n -1+2 and bn=an+bn-1 for every natural number n ≥ 2. Then displaystyle∑n=115 an ⋅ bn is equal to  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

a 1= b 1=1 a 2= a 1+2=3 a 3= a 2+2=5 a 4= a 2+2=7 ⇒ a n =2 n -1 b 2= a 1+ b 1=4 b 3= a 3+ b 2=9 b 4= a 4+ b 3=16 b n = n 2 displaystyle∑n=115 an bn displaystyle∑n=115(2 n-1) n2 displaystyle∑n=115(2 n3-n2) =2 (n2(n+1)2/4)-(n(n+1)(2 n+1)/6) Put n=15 =(2 × 225 × 16 × 16/4)-(15 × 16 × 31/6)=27560

Q. Let a1​=b1​=1,an​=an​−1+2 and bn​=an​+bn−1​ for every natural number n≥2. Then n=1∑15​an​⋅bn​ is equal to ______

a1​=b1​=1 a2​=a1​+2=3 a3​=a2​+2=5 a4​=a2​+2=7 ⇒an​=2n−1 b2​=a1​+b1​=4 b3​=a3​+b2​=9 b4​=a4​+b3​=16 bn​=n2 n=1∑15​an​bn​ n=1∑15​(2n−1)n2 n=1∑15​(2n3−n2) =24n2(n+1)2​−6n(n+1)(2n+1)​ Put n=15 =42×225×16×16​−615×16×31​=27560

Q. Let $a_{1} = b_{1} = 1, a_{n} = a_{n} - 1 + 2$ and $b_{n} = a_{n} + b_{n - 1}$ for every natural number $n \geq 2$ .
Then $n = 1 \sum 15 a_{n} \cdot b_{n}$ is equal to ______