Let \langle a_n\rangle be a sequence such that a_0=0, a_1=\frac12 and 2 a_n+2=5 a_n+1-3 a_n, n=0,1,2,3, \ldots . Then \sum\limits_k=1^100 a_k is equal to

Sequences and Series PYQ — JEE Main 2025 · 28 Jan · Shift 1 · Q30 with Solution

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Correct answer: B

We have $a_0=0,\, a_1=\frac12$ and $2a_{n+2}=5a_{n+1}-3a_n \qquad (n\ge 0).$

We need to find $\sum_{k=1}^{100} a_k.$

Rewrite the recurrence as $2a_{n+2}-5a_{n+1}+3a_n=0.$

The characteristic equation is $2r^2-5r+3=0.$

Factorizing, $2r^2-5r+3=(2r-3)(r-1)=0.$

So the roots are $r=1,\quad r=\frac32.$

Hence the general term is $a_n=A+B\left(\frac32\right)^n.$

From $a_0=0$ , $A+B=0 \implies A=-B.$

From $a_1=\frac12$ , $A+\frac32 B=\frac12.$

Substitute $A=-B$ : $-B+\frac32 B=\frac12$ $\frac12 B=\frac12 \implies B=1.$

Thus, $A=-1.$

Therefore, $a_n=\left(\frac32\right)^n-1.$

Now,

=\sum_{k=1}^{100}\left(\frac32\right)^k-100.$$ The geometric sum is $$\sum_{k=1}^{100}\left(\frac32\right)^k =\frac{\frac32\left[\left(\frac32\right)^{100}-1\right]}{\frac32-1}.$$ Since $\frac32-1=\frac12$, $$\sum_{k=1}^{100}\left(\frac32\right)^k =3\left[\left(\frac32\right)^{100}-1\right].$$ So, $$\sum_{k=1}^{100} a_k=3\left[\left(\frac32\right)^{100}-1\right]-100.$$ But $$a_{100}=\left(\frac32\right)^{100}-1.$$ Hence, $$\sum_{k=1}^{100} a_k=3a_{100}-100.$$ --- 5. **Check options** - A: $3a_{100}+100$ ❌ - B: $3a_{100}-100$ ✅ - C: $3a_{99}-100$ ❌ - D: $3a_{99}+100$ ❌ So the correct option is **B**.